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THEORETICAL    MECHANICS 


STANFORD  UNIVERSITY   BOOKSTORE 
Agent  of  the  Publisher 


Theoretical  Mechanics 


%n  (&\tmtntM\j  Stxt-tarok 


BY 


L.   M.   HOSKINS 


Professor  of  Applied  Mathematics  in  the 
Leland  Stanford  Junior  University 


SECOND  EDITION 


' 


STANFORD  UNIVERSITY,  CAL. 

PUBLISHED  BY  THE   AUTHOR 
1903 


HALLIDIE  ^-^ 


Copyright,  1900,  1903, 
By  L.  M.  Hoskins 


Stanford  University  Press 


PREFACE. 


In  a  first  course  in  Theoretical  Mechanics  the  primary  object  to 
be  gained  by  the  student  is  a  thorough  grasp  of  fundamental  princi- 
ples. In  most  cases  it  is  impossible  to  go  beyond  this  object  in  the 
time  available  for  the  course.  In  the  preparation  of  this  text-book, 
the  aim  has  been  to  present  the  fundamental  principles  in  as  clear 
and  simple  a  manner  as  possible,  and  to  enforce  them  by  a  sufficient 
number  of  illustrative  examples.  The  examples  are  for  the  most 
part  simple  applications  of  the  theory  presented  in  the  text,  many 
numerical  exercises  being  included.  The  solution  of  exercises  in- 
volving numerical  data  forms  a  part  of  the  work  of  the  student  of 
which  the  importance  should  be  emphasized.  In  the  desire  to  cover 
as  much  ground  as  possible,  such  work  is  too  apt  to  be  neglected. 

The  mathematical  training  required  for  using  the  book  is  that 
usually  implied  by  an  elementary  knowledge  of  Differential  and  Inte- 
gral Calculus.  The  most  essential  portions  of  Part  I  (Statics)  may, 
however,  be  read  without  such  knowledge. 

As  regards  arrangement,  the  aim  has  been  to  observe  the  order 
of  difficulty  of  the  different  topics  treated,  so  far  as  this  was  possible 
without  doing  violence  to  the  logical  relations.  This  is  one  reason 
for  placing  Statics  first,  instead  of  treating  it  in  its  strictly  logical 
position  as  a  special  case  of  Kinetics.  Part  II  may,  however,  be 
read  before  Part  I,  if  any  teacher  should  prefer  this  order. 

As  to  scope,  the  primary  object  has  been  to  meet  the  needs  of 
students  of  engineering  in  American  universities  and  technical  col- 
leges. More  has,  however,  been  included  than  it  will  usually  be 
possible  or  desirable  to  cover  in  an  elementary  course.  The  precise 
limits  of  the  course  must  be  determined  by  the  judgment  of  the 
teacher,  taking  due  account  of  the  relative  importance  of  the  various 
parts  of  the  subject  and  of  the  limitations  of  time.  It  is  hoped  that 
the  arrangement  of  the  book  is  such  as  to  facilitate  whatever  abridg- 
ment the  teacher  may  find  desirable. 

In  case  it  is  necessary  to  cut  the  course  down  to  the  narrowest 
possible  limits,  the  following  suggestions  as  to  the  most  essential 
parts  to  retain  may  be  of  service. 

117270 


VI  PREFACE. 

Chapters  I-V.  These  include,  besides  the  introductory  general 
matter,  the  fundamental  principles  governing  the  composition  and 
resolution  of  forces  and  the  conditions  of  equilibrium,  as  applied  to 
a  rigid  body  acted  upon  by  coplanar  forces. 

Chapters  XII-XVII  ;  omitting  some  of  the  applications  in  Chap- 
ters XIII  and  XV,  and  §  2  of  Chapter  XVI.  These  chapters  cover 
the  most  essential  parts  of  the  theory  of  the  motion  of  a  particle, 
including  a  general  discussion  of  the  laws  of  motion  and  an  introduc- 
tion to  the  theory  of  energy. 

Chapters  VI  and  VII  are  of  practical  value  to  the  student  of  en- 
gineering, and  the  course  should  include  at  least  §  1  and  §  2  of 
Chapter  IX. 

Chapters  XVIII-XX  constitute  an  introduction  to  the  Dynamics 
of  Material  Systems,  with  applications  to  the  rotation  of  a  rigid  body. 
If  possible,  at  least  this  portion  of  Part  III  should  be  included  in 
the  course.  In  Chapter  XIX,  §  2  (methods  of  computing  moments 
of  inertia  of  plane  areas)  is  of  practical  value  for  its  application  in 
Mechanics  of  Materials,  and  may  be  taken  independently  of  any  other 
portion  of  Part  III. 

If  this  book  is  in  any  degree  successful  in  meeting  the  needs  of 
students  of  engineering,  it  is  hoped  that  it  may  be  of  service  also  to 
those  pursuing  the  subject  for  its  intrinsic  scientific  interest  or  as  a 
preparation  for  the  study  of  mathematical  physics.  The  opinion  is 
sometimes  expressed  that  the  needs  of  these  different  classes  of  stu- 
dents require  essentially  different  methods  of  treating  the  subject. 
This  view,  so  far  as  it  refers  to  the  fundamental  parts  of  an  elementary 
course,  is  not  shared  by  the  author  of  this  text-book.  For  all  stu- 
dents, the  matter  of  first  importance  is  the  clear  understanding  of 
fundamental  general  principles  and  the  ability  to  apply  them. 

Stanford  University,  Cal.,  October,  1900. 


PREFACE  TO   SECOND   EDITION. 


In  this  edition  Chapters  I-XXIII  are  reprinted  from  the  first 
edition  without  important  change,  except  the  correction  of  such 
typographical  and  other  errors  as  have  been  detected.  Chapter 
XXIV  is  new. 

Stanford  University,  Cal.,  August,  1903. 


CONTENTS. 


CHAPTER 

I.  Introductory. 


Preliminary  Notions ;    Numerical   Representation  of  Quanti- 
ties ;  Vector  Quantities. 


PART   I.     STATICS. 

II.  Force  and  Stress.  .  .  .  .  .  .16 

Conception  of  Force ;  Classes  of  Forces ;  Numerical  Repre- 
sentation of  Forces  and  of  Masses ;  Definitions. 

III.  Concurrent  Forces.         .  .  .  .  .  .27 

Composition  and  Resolution;  Moments;  Equilibrium. 

IV.  Composition  and  Resolution  oe  Non-concurrent  Forces 

in  the  Same  Plane.  .  .  .  .  .49 

Two  Forces;  Couples;  Any  Number  of  Forces. 

V.  Equilibrium  of  Coplanar  Forces.        .  .  .  -65 

General  Principles ;  Applications. 

VI.  Equilibrium  of  Parts  of  Bodies  and  of  Systems  of  Bodies.      92 
Equilibrium  of  Any  Part  of  a  Body  ;   Determination  of  Inter- 
nal Forces ;   Equilibrium  of  a  System  of  Bodies  ;   Stress. 

VII.  Friction.        .  .  »  .  .  .  .  .     106 

VIII.  Equilibrium  of  Flexible  Cords.  .  .  .  .114 

IX.  Centroids.     .  .  .  .  .  .  .  .124 

Centroid  of  Parallel  Forces;   Centroids  of  Masses,  Volumes, 
Areas  and  Lines  ;  Determination  of  Centroids  by  Integration. 

X.  Forces  in  Three  Dimensions.  .  .  .  .141 

Concurrent  Forces ;  Couples  ;  Non-concurrent  Forces ;  Equi- 
librium. 

XI.  Gravitation.  .  .  .  .  .  .  .     155 

Attraction  Between  Particles ;   Attractions  of  Spheres  and  of 
Spherical  Shells. 


CONTENTS. 


PART  II.     MOTION  OF  A  PARTICLE. 

CHAPTER  PAGE 

XII.  Motion  in  a  Straight  Line:   Fundamental  Principles.    .     167 
Position,    Displacement    and    Velocity;    Velocity-Increment 
and  Acceleration ;   Motion  and  Force. 

XIII.  Motion  in  a  Straight  Line:   Applications.  .  .     186 

General  Method ;  Constant  Force ;  Force  Varying  with  Dis- 
tance from  Fixed  Point ;  Miscellaneous  Problems. 

XIV.  Motion  in  a  Curved  Path.  .  .  .  .     208 

Position,  Displacement  and  Velocity ;  Velocity-Increment  and 
Acceleration;  Motion  and  Force ;  Simultaneous  Motions. 

XV.  Plane  Motion  of  a  Particle.  .  .  .  .236 

Methods  of  Specifying  Motion  in  a  Plane ;  Motion  Under  Any 
Forces;  Resultant  Force  Constant  or  Zero;  Central  Force; 
Constrained  Motion. 

XVI.  Momentum  and  Impulse.  .  .  .  .  .277 

Rectilinear  Motion;  Motion  in  Any  Path. 

XVII.  Work  and  Energy.  .  .  .  .  .  .298 

Work  in  Case  of  Rectilinear  Motion ;  Work  in  Any  Motion ; 
Energy  of  a  Particle ;  Energy  of  a  System ;  Virtual  Work. 


PART  III.     MOTION  OF  SYSTEMS  OF  PARTICLES  AND 
OF   RIGID   BODIES. 

XVIII.  Motion  of  Any  System  of  Particles.  .  .  .321 

Motions  of  Individual  Particles  and  of  Center  of  Mass;  An- 
gular Motion ;   Effective  Forces ;   D'Alembert's  Principle. 

XIX.  Moment  of  Inertia.        ......     331 

Rigid  Body ;   Plane  Area. 

XX.  Motion  of  a  Rigid  Body:  Translation,  Rotation  About 

a  Fixed  Axis.  ......     346 

XXI.  Any  Plane  Motion  of  a  Rigid  Body.  .  .  .     360 

Nature  of    Plane   Motion;    Composition   and    Resolution  of 
Plane  Motions ;   Dynamics  of  Plane  Motion ;  Statics. 


CONTENTS.  XI 

CHAPTER  PAGE 

XXII.  Principle  of  Impulse  and  Momentum.         .  .  .      382 

Any  System  of  Particles  ;  Rigid  Body  Having  Motion  of 
Translation  or  of  Rotation ;  Resultant  Momentum  ;  Any  Plane 
Motion  of  Rigid  Body. 

XXIII.  Theory  of  Energy.       ......      405 

External  and  Internal  Work  ;  Energy  of  Any  System  of  Par- 
ticles ;  Conservation  of  Energy ;  Rigid  System  ;  Principle  of 
Virtual  Work. 

XXIV.  Relative  Motion.  ......     432 

Index.  .........      453 


THEORETICAL  MECHANICS 


OF  TH-.. 


CHAPTER  I. 

INTRODUCTORY. 

§  i.   Preliminary  Notions, 


i.  Mechanics  Defined. —  Mechanics  is  the  science  which  treats 
of  the  motions  of  material  bodies. 

Motions  take  place  in  accordance  with  definite  laws.  This  state- 
ment means  that  the  motions  of  bodies  depend  in  an  invariable  way 
upon  certain  definite  conditions.  The  fundamental  object  of  the 
science  of  Mechanics  is  to  investigate  these  conditions  and  to  formulate 
the  laws  in  accordance  with  which  they  determine  the  motion. 

In  its  present  form  the  science  of  Mechanics  rests  upon  a  few 
fundamental  laws  of  great  generality.  These  laws  are  generaliza- 
tions from  experience,  and  in  the  presentation  of  the  principles  of 
the  science  they  are  taken  as  postulates, —  propositions  not  deduc- 
ible  from  anything  more  fundamental.  So  far  as  these  postulates 
and  the  principles  deduced  from  them  give  a  true  account  of  the 
motions  of  natural  bodies,  they  constitute  a  natural  science. 

These  fundamental  laws,  however,  involve  certain  conceptions  of 
matter  and  of  motion  which  are  ideal.  This  is  necessarily  the  case. 
The  motion  of  a  body  can  be  completely  specified  only  by  describ- 
ing the  motion  of  every  ultimate  portion  of  which  it  is  composed  ; 
yet  the  ultimate  structure  of  matter  is  wholly  unknown.  The  bodies 
to  which  the  laws  of  motion  apply  are  therefore  defined  in  an  ideal 
way.  Moreover,  the  very  conception  of  motion  involves  the  abstract 
notions  of  Geometry,  with  the  notion  of  time  added.  Because  of 
this  ideal  character  of  the  laws,  the  science  based  upon  them  is 
properly  called  Theoretical  Mechanics. 


2  THEORETICAL   MECHANICS. 

2.  Material  Bodies. —  A  body  is  any  definite  portion  of  matter. 
No  attempt  need  be  made  to  define  matter  or  to  enumerate  its 

properties.     A  sufficiently  definite  preliminary  notion  is  supplied  by 
ordinary  experience. 

The  characteristics  of  material  bodies  which  are  of  importance  in 
a  study  of  their  motions  are  the  following  :  (i)  Every  body  has  a 
definite  volume  and  a  definite  figure ;  (2)  every  body  possesses  a 
definite  mass  ;  (3)  bodies  exert  forces  upon  one  another. 

3.  Mass. —  The  mass  of  a  body  is  often  briefly  defined  as  its 
"quantity  of  matter."  These  words,  however,  convey  no  definite 
idea  of  the  meaning  of  mass  as  a  factor  in  the  determination  of 
motion.  A  satisfactory  definition  of  mass  cannot  be  given  in 
advance  of  a  discussion  of  the  fundamental  laws  of  motion. 

We  conceive  of  mass  as  the  one  invariable  characteristic  of  mat- 
ter. Every  individual  portion  of  matter  is  regarded  as  possessing 
a  definite  mass  whose  value  is  uninfluenced  by  changes  of  position  or 
by  physical  or  chemical  transformations.  The  volume  and  shape  of 
a  body  may  change,  the  forces  it  exerts  upon  other  bodies  and  those 
which  they  exert  upon  it  are  different  under  different  circumstances  ; 
but  its  mass  is  regarded  as  an  absolute  constant. 

4.  Force. —  A  force  is  an  action  exerted  by  one  body  upon 
another,  tending  to  change  the  state  of  motion  of  the  body  acted 
upon. 

A  force  may  be  conceived  as  a  push  or  a  pull,  acting  upon  a 
definite  portion  of  a  body.  Such  a  push  or  pull  always  tends  to 
change  the  motion  of  the  body  ;  but  this  tendency  may  be  counter- 
acted in  whole  or  in  part  by  the  action  of  other  forces. 

Mechanics  is  often  called  the  science  of  motion  and  force,  because 
of  the  importance  of  force  in  the  development  of  the  laws  of  the 
science. 

Force,  like  mass,  is  a  quantity  whose  significance  cannot  be  satis- 
factorily explained  except  by  a  full  discussion  of  the  fundamental  laws 
of  motion. 

5.  Particle. —  A  body  may  be  conceived  to  be  divided  into  very 
small  parts,  each  of  which  may  be  called  a  particle.  Ideally,  there 
is  no  limit  to  this  process  of  subdivision.  For  the  purposes  of 
mathematical  analysis  it  is  often  conceived  to  be  carried  so  far  that 
the  linear  dimensions  of  the  particles  become  vanishingly  small. 


INTRODUCTORY.  3 

These  particles  may  be  conceived  in  either  of  two  ways,  corre- 
sponding to  two  different  conceptions  of  the  structure  of  matter. 

(1)  It  may  be  assumed  that  matter  occupies  space  continuously. 
By  this  it  is  to  be  understood  that  every  portion  of  matter  whose 
mass  is  finite  occupies  a  finite  volume. 

(2)  It  may  be  assumed  that  any  definite  portion  of  matter  con- 
sists of  particles,  each  of  which  possesses  finite  mass  but  occupies  no 
finite  volume. 

The  hypothesis  of  continuity  does  not  necessarily  imply  that  there 
may  not  be  void  spaces  between  the  parts  of  a  body.  In  mathe- 
matical language  its  meaning  may  be  stated  as  follows  :  Let  M  be 
the  mass  of  a  body  and  V  its  volume  ;  and  let  bkM  be  the  mass  of 
a  small  portion  whose  volume  is  A  V.  Then  if  A  V  is  taken  smaller 
and  smaller  so  that  A  Vj  V  approaches  zero,  the  hypothesis  of 
continuity  implies  that  AM/M  also  approaches  zero  ;  while  the 
hypothesis  of  discontinuity  implies  that  AM/M  may  be  finite. 

In  certain  discussions  it  is  found  convenient  to  adopt  one  hypoth- 
esis, in  other  cases  the  other  ;  so  far  as  results  are  concerned,  it  is 
usually  immaterial  which  is  adopted. 

In  the  analysis  of  the  motion  of  a  particle,  it  is  regarded  as  a 
geometrical  point  endowed  with  mass.  A  body  whose  linear  dimen- 
sions are  small  in  comparison  with  the  range  of  its  motion  is  often 
regarded  as  a  particle. 

6.  Rigid  Body. —  A  rigid  body  may  be  defined  as  a  body  whose 
particles  do  not  change  their  distances  from  one  another. 

An  important  part  of  the  science  of  Theoretical  Mechanics  deals 
with  bodies  which  are  assumed  to  satisfy  this  definition. 

Actual  solid  bodies  undergo  appreciable  changes  of  shape  and 
size  ;  and  if  sufficiently  small  portions  could  be  observed,  they  would 
doubtless  be  found  to  be  in  rapid  motion,  thus  departing  very  far 
from  the  condition  specified  in  the  definition  of  rigidity.  But  dis- 
regarding the  motions  of  ultimate  particles  and  considering  only  the 
motion  of  a  body  as  a  whole,  the  theory  of  the  motion  of  an  ideal 
rigid  body  describes  with  great  accuracy  the  motion  of  an  actual  solid 
body  whose  shape  remains  nearly  constant. 

7.  Position  and  Motion. —  In  the  foregoing  definitions  and  ex- 
planations it  has  been  assumed  that  position  and  motion  need  no 
definitions.  It  is,  in  fact,  doubtful  whether  any  definitions  can  be 
given  which  convey  clearer  notions  than  the  words  themselves. 


4  THEORETICAL    MECHANICS. 

By  the  position  of  a  particle  is  meant  its  relation  in  space  to  some 
body  taken  as  a  standard  of  reference. 

A  particle  is  in  motion  when  its  position  is  changing.  A  body  is 
in  motion  when  the  particles  composing  it  are  in  motion. 

What  shall  be  chosen  as  the  body  of  reference  in  specifying  the 
position  and  motion  of  a  particle  is  a  matter  of  arbitrary  choice. 
Position  and  motion  are  thus  not  absolute  but  relative.  The  motion 
of  a  particle  with  reference  to  one  body  may  be  very  different  from 
its  motion  with  reference  to  another.     (See  Arts.  267,  268.) 

8.  Kinds  of  Quantity. —  The  science  of  Mechanics  deals  with 
quantities  of  four  fundamental  kinds  :  time,  space,  mass,  force.  In 
the  development  of  the  principles'of  the  science  other  quantities  are 
introduced  which  are  derived  from  these  but  involve  no  other 
elementary  conceptions.* 

9.  Divisions  of  the  Subject. —  The  general  subject  of  Mechanics 
may  be  subdivided  in  various  ways. 

First,  the  basis  of  the  subdivision  may  be  the  nature  of  the  bodies 
dealt  with.     On  this  basis  there  are  the  following  divisions  : 

(a)  Mechanics  of  a  Particle  and  of  systems  of  particles  in  general. 

(b)  Mechanics  of  Solid  Bodies \  including  (1)  rigid  and  (2)  non- 
rigid  solids. 

if)  Mechanics  of  Fluids,  including  (1)  liquids  and  (2)  gases. 

The  present  work  deals  mainly  with  particles  and  with  rigid  solids. 

Second,  the  basis  of  subdivision  may  be  the  fundamental  kinds 
of  quantity  involved.  On  this  basis  the  whole  subject,  and  each  of 
the  above  divisions,  may  be  divided  as  follows  : 

{a)  Kinematics,  treating  of  motion,  without  reference  to  the  causes 
producing  or  influencing  it.  Under  this  there  are  (1)  Pure  Kine- 
matics, treating  of  motion  apart  from  the  idea  of  mass,  and  (2) 
Mass-Kinematics,  treating  of  motion  and  mass. 

(b)  Dynamics,  treating  of  forces  and  of  their  influence  upon  the 
motions  of  bodies.  Dynamics,  the  science  of  force,  is  again  sub- 
divided into  Statics  and  Kinetics. 

*  It  is  pointed  out  in  a  later  chapter  that  these  four  quantities  are  not 
always  regarded  as  being  all  fundamental.  In  particular,  force  is  often 
regarded  as  a  quantity  which  must  be  defined  in  terms  of  mass,  space  and 
time.  Ordinary  experience,  however,  supplies  a  tolerably  definite  notion  of 
force  which  is  wholly  independent  of  any  conception  of  the  meaning  of  the 
laws  of  motion.    See  "  The  Six  Gateways  of  Knowledge,"  by  Lord  Kelvin. 


INTRODUCTORY.  5 

Statics  treats  of  the  conditions  of  equivalence  of  systems  of  forces, 
and  especially  of  the  conditions  under  which  forces  are  balanced 
so  that  they  do  not  affect  the  motions  of  the  bodies  acted  upon. 
Kinetics  treats  of  the  laws  in  accordance  with  which  the  motions  of 
bodies  are  influenced  by  the  forces  acting  upon  them  and  by  their 
masses. 

Strictly  speaking,  Kinetics  includes  Statics  ;  for  from  a  knowledge 
of  the  effects  of  forces  upon  the  motions  of  bodies  may  be  derived  all 
principles  relating  to  the  equivalence  of  different  systems  of  forces 
and  to  the  conditions  under  which  forces  are  balanced.  Statics  and 
Kinetics  are,  however,  often  treated  as  coordinate  branches  of  Dy- 
namics ;  first,  because  Statics,  although  a  special  case,  is  a  case  of 
great  importance,  and  second,  because  the  principles  of  Statics  can 
be  developed  to  a  large  extent  independently  of  those  of  Kinetics. 

The  arrangement  of  subjects  in  this  book  does  not  strictly  follow 
either  of  the  above  classifications,  but  is  designed  to  meet  the  needs 
of  beginners  by  presenting  the  different  topics  somewhat  in  the  order 
of  their  difficulty.     The  arrangement  adopted  is  as  follows  : 

Introductory  chapter,  treating  of  certain  principles  which  have 
application  in  various  branches  of  the  general  subject  of  Mechanics. 

Part  I.  Statics  ;  the  discussion  of  the  subject  being  limited  mainly 
to  systems  of  forces  acting  in  the  same  plane  upon  a  particle  or  a 
rigid  body. 

Part  II.  Motion  of  a  Particle  ;  including  Kinematics  and  Kinetics, 
but  limited  mainly  to  the  case  of  motion  in  a  plane. 

Part  III.  Motion  of  Systems  of  Particles  and  of  Rigid  Bodies  ; 
including  Kinematics  and  Kinetics,  but  dealing  mainly  with  motion 
in  a  plane. 

The  remaining  portion  of  this  introductory  chapter  is  devoted  to 
certain  principles  which,  while  not  strictly  included  in  the  subject  of 
Mechanics,  are  of  fundamental  importance  in  the  development  of  the 
science. 

§  2.    The  Numerical  Representation  of  Quantities. 

10.  Comparison  of  Like  Quantities. —  The  magnitudes  of  any 
two  quantities  of  the  same  kind  (as  two  distances,  or  two  intervals 
of  time,  or  two  forces)  may  be  compared  by  determining  their  ratio. 
Such  a  ratio  is  an  abstract  number. 

11.  Numerical  Value  of  a  Quantity. —  The  numerical  value  of 


6  THEORETICAL    MECHANICS. 

a  quantity  is  the  ratio  of  its  magnitude  to  that  of  a  given  quantity  of 
the  same  kind,  taken  as  a  standard.  This  standard  quantity  is  called 
a  unit.  Thus,  to  express  a  length  as  a  number,  some  definite  length 
must  be  chosen  as  a  unit  (its  numerical  value  being  called  one)  ; 
then  the  numerical  value  of  any  given  length  will  be  the  number 
expressing  its  ratio  to  the  unit  length. 

The  unit  in  terms  of  which  quantities  of  a  given  kind  are  ex- 
pressed is  wholly  arbitrary.  Thus,  there  are  in  common  use  several 
different  units  of  length,  as  the  foot,  the  inch,  the  yard,  the  mile,  the 
meter  ;  several  units  of  time,  as  the  second,  the  minute,  the  week, 
the  year ;  and  several  units  of  mass,  as  the  pound,  the  ton,  the 
ounce,  the  gram,  the  kilogram. 

12.  Numerical  Value  Depends  on  Unit. —  Evidently,  the  nu- 
merical value  of  a  quantity  gives  no  idea  of  its  magnitude  unless  the 
unit  employed  is  known.  Any  given  quantity  may  be  represented 
by  any  number  whatever,  by  properly  choosing  the  unit.  Thus, 
the  same  distance  may  be  called  10,560  feet,  3,520  yards,  or  2  miles. 

But  the  ratio  of  two  quantities  of  the  same  kind  is  a  definite 
number,  which  depends  only  upon  the  magnitudes  of  the  quantities, 
and  not  upon  their  numerical  values.  The  ratio  of  the  numerical 
values  of  two  quantities  is  the  same  as  the  ratio  of  the  quantities 
themselves  if  they  are  expressed  in  the  same  unit,  but  not  otherwise. 

13.  Fundamental  and  Derived  Units. —  Although,  as  above 
stated,  the  unit  in  terms  of  which  quantities  of  any  particular  kind 
are  expressed  numerically  may  be  chosen  arbitrarily,  it  is  usually 
advantageous  to  choose  the  units  for  quantities  of  different  kinds  in 
such  a  way  that  certain  of  them  depend  upon  others.  In  Physics, 
the  usual  practice  is  to  choose  arbitrarily  the  units  of  time,  length 
and  mass,  and  to  make  all  other  units  depend  upon  these.  Those 
units  which  are  chosen  arbitrarily  are  called  fundamental,  while  those 
which  are  so  defined  as  to  depend  upon  the  fundamental  units  are 
called  derived  units. 

For  example,  the  unit  of  area  is  usually  taken  as  the  area  of  a 
square  whose  side  has  the  unit  length.  It  should,  however,  be 
clearly  understood  that  this  dependence  is  not  necessary,  but  is 
merely  assumed  for  convenience. 

14.  Dimensions  of  Derived  Units. —  The  word  dimension  is 
commonly  employed  to  describe  the  way  in  which  a  derived  unit 
depends  upon  the  fundamental  units. 


INTRODUCTORY.  7 

Thus,  if  the  unit  area  is  denned  as  the  area  of  a  square  whose  side 
has  the  unit  length,  the  relation  may  be  expressed  by  the  "dimen- 
sional equation ' ' 

(unit  area)  =  (unit  length) 2, 

and  the  unit  area  is  said  to  involve  two  dimensions  of  length.  Sim- 
ilarly, if  the  unit  volume  is  denned  as  the  volume  of  a  cube  whose 
edge  has  the  unit  length,  the  unit  volume  involves  three  dimensions 
of  length. 

In  the  illustrations  just  given,  only  one  kind  of  fundamental 
quantity  (length)  is  employed.  There  will  be  occasion  in  the  follow- 
ing pages  to  deal  with  derived  units  which  depend  upon  two  or 
more  fundamental  units. 

Evidently  the  manner  in  which  a  derived  unit  is  made  to  depend 
upon  the  fundamental  units  is  to  some  extent  arbitrary,  and  by 
changing  the  manner  of  dependence  the  dimensions  of  any  given 
kind  of  quantity  may  be  changed.  This  subject  will  be  further 
treated  in  connection  with  the  various  kinds  of  quantity  dealt  with 
in  the  following  pages. 

15.  Dimensional  Equations. —  The  dimensional  equation  above 
written  is  not  to  be  interpreted  as  an  ordinary  algebraic  equation. 
It  is  merely  a  concise  method  of  expressing  the  way  in  which  the 
derived  unit  (of  area)  is  made  to  depend  upon  the  fundamental  unit 
(of  length).  It  will  be  convenient  to  abbreviate  such  equations  by 
the  use  of  symbols  to  denote  the  various  units.  The  quantities 
whose  units  are  taken  as  fundamental  in  the  following  pages  are 
usually  some  or  all  of  the  following  :  length,  mass,  time,  force.  In 
writing  dimensional  equations  the  letters  L,  M,  T,  F  will  be  used  to 
designate  the  units  of  these  four  kinds  of  quantity. 


§  3.    Vector  Quantities. 

16.  Vector  Quantity  Defined. —  A  directed  straight  fine  of  def- 
inite length  is  called  a  vector. 

A  vector  quantity  is  one  possessing  direction  as  well  as  mag- 
nitude. 

Any  vector  quantity  may  be  represented  by  a  vector.  For  this 
purpose  the  direction  of  the  vector  must  agree  with  that  of  the  quan- 


8  THEORETICAL    MECHANICS. 

tity  to  be  represented,  and  its  length  must  represent  the  magnitude 
of  the  quantity. 

In  order  that  the  length  of  a  line  may  represent  the  magnitude 
of  any  given  quantity,  a  scale  must  be  chosen.  That  is,  a  certain 
length  must  be  taken  as  equivalent  to  a  unit  magnitude  of  the  kind 
to  be  represented.  The  length  thus  chosen  is  arbitrary,  but  when 
once  chosen,  the  length  representing  any  definite  magnitude  of  the 
kind  in  question  is  determined. 

A  vector  may  be  designated  by  naming,  in  proper  order,  the 
letters  placed  at  its  ends.  Thus,  if  A  and  B  are  any  two  points, 
AB  and  BA  denote  two  vectors  of  equal  length  but  opposite  in  di- 
rection. 

A  vector  may  also  be  designated  by  a  single  symbol,  as  in  the 
case  of  ordinary  algebraic  quantities. 

Scalar. —  The  processes  of  ordinary  algebra  are  limited  to  quan- 
tities which  either  do  not  involve  direction  or  else  are  restricted  to 
two  opposite  directions.  Such  quantities  are  called  scalars.  A 
scalar  quantity  is  completely  specified  by  assigning  its  magnitude  and 
algebraic  sign. 

17.  Equality  of  Vectors. —  Two  vectors  are  said  to  be  equal  if 
they  agree  not  only  in  magnitude  but  in  direction.     Thus,  if  MN 

and  PQ  (Fig.  1)  are  parallel  and  of  equal 
length,  we  may  write 

MN  =  PQ. 

If  this  were  an  ordinary  algebraic  equation, 
it  would  express  only  equality  of  length, 
and  would  be  equally  true  if  the  lines  were 
not  parallel  ;  but  as  a  vector  equation  it 
expresses  identity  of  direction  as  well  as 
equality  of  length.     Thus,  referring  to  Fig.  1, 

MN=QP 

is  not  true  as  a  vector  equation. 

18.  Addition  of  Vectors. —  Non-parallel  vectors  cannot  be  added 
in  the  sense  in  which  scalars  are  added  ;  but  the  word  addition  is 
used  to  describe  the  process  of  combining  two  vectors  in  the  follow- 
ing manner  : 


INTRODUCTORY.  Q 

Let  MN  and  PQ  (Fig.  2)  represent  two  vectors  ;  then  by  their 
sum  is  meant  a  vector  determined  as  follows  :  Make  AB  equal  and 
parallel  to  MN,  and  BC  equal  and  par- 
allel to  PQ  ;  then  A  C  represents  the  re- 
quired sum.     In  fact,  there  may  be  writ- 
ten the  equation 

AB  +  BC  =  AC;  or  MN  +  PQ  =  A C. 

But  this  must  be  understood  as  a  vec- 
tor equation,  and  not  as  applying  to  the 
lengths  MN,  PQ  and  A  C 

If  the  order  of  the  two  vectors  is 
changed  their  sum  remains  the  same. 
For  if  lines  equal  and  parallel  to  PQ  and 
MN  be  laid  off  successively  from  A ,  the 
result  will  be  AB'  and  B' C,  which,  with 
AB   and  BC,   form  a  parallelogram  of  which  AC  is  a   diagonal. 

Next,  let  any  three  vectors  be  drawn  consecutively,  as  AB,  BC, 
CD  (Fig.  3)  ;  then  AD  is  called  their  sum.  It  is  easily  seen  that 
the  three  vectors  may  be  drawn  in  any  order  without  changing  their 

sum  as  thus  defined.      It  is  also  evident 
B  ^]D     that 

AB  +  BC+  CD  =  AC  +  CD 
=  (AB  +  BC)  +  CD ; 

that  is,   the  result  of  adding  the  three 
Fig.  3.  vectors  is  the  same  as  the  result  of  add- 

ing any  one  to  the  sum  of  the  other  two. 
The  same  construction  may  be  extended  to  any  number  of  vec- 
tors. 

19.  Vector  Subtraction. —  Parallel  vectors  may  be  distinguished 
in  direction  by  signs  plus  and  minus  as  in  Algebra.  Thus,  if  A  and 
B  are  any  two  points,  the  two  vectors  AB  and  BA  may  be  desig- 
nated by  prefixing  signs  plus  and  minus  to  the  same  symbol ;  if  the 
symbol  p  represents  the  vector  AB,  — /represents  the  vector  BA. 

The  minus  sign  may  thus  be  used  to  denote  reversal  of  the  di- 
rection of  a  vector,  so  that  we  may  write 


AB 


BA. 


IO 


THEORETICAL    MECHANICS. 


It  may  also  be  used  to  denote  subtraction,  in  accordance  with  the 
following  definition  : 

To  subtract  a  vector  is  to  add  its  negative. 

With  this  definition  of  subtraction,  the  difference  between  two 
vectors  may  be  found  as  follows  : 

Let  it  be  required  to  subtract  the  vector  MN  from  the  vector 
PQ  (Fig.  4).     Lay  off  AB  =  MN  and 
AC  =  PQ;  then 

PQ~  MN       AC—  AB 
=  AC+BA  =BA  +  AC 
=  BC 

20.  Vector  Equations. —  From  the 
above  definitions  of  vector  addition  and 
subtraction,  vector  equations  may  be 
formed  containing  any  number  of  terms 

in  each  member,  and  may  be  transformed  in   accordance  with  sim- 
ple rules.     Thus,  if  A,  B,  C,  D,  E,  F  are  any  six  points,  we  may 

write  the  vector  equa- 
B  tion 

AB  +  BC+CD  = 

AE  +  EF+FD;    (i) 

for  each  member  of  the 
m     equation  is  equal  to  the 
vector  AD. 

In  several    particu- 
lars   vector     equations 
may  be  treated   by  the 
same  rules  as  the  equations  of  ordinary  Algebra. 

( i )  Equal  vectors  may  be  added  to  each  member  of  a  vector  equa- 
tion without  destroying  the  equality.  Thus,  if  a  vector  DM  be 
added  to  each  member  of  equation  (i),  each  member  becomes  equal 
to  AM,  and  therefore  the  equality  still  holds.  By  an  extension  of 
this  principle,  any  number  of  vector  equations  may  be  added,  mem- 
ber to  member. 

(2)  A  term  may  be  transposed  from  one  member  to  the  other  by 
changing  its  sign.  Thus,  from  equation  (1)  we  may  derive  the 
equation 


INTRODUCTORY.  II 

AB  +  BC  =  AE  +  EF+FD  +  DC, 

or  AB  +  BC=AE  +  EF -\- FD  ~  CD, 

by  transposing  the  term  CD  and  changing  its  sign. 

(3)  The  order  of  the  terms  in  either  member  of  a  vector  equa- 
tion may  be  changed  at  pleasure.  For,  as  already  pointed  out, 
changing  the  order  in  which  several  vectors  are  combined  does  not 
change  their  sum. 

21.  Constant  and  Variable  Vector  Quantities. —  A  vector  quan- 
tity may  be  either  constant  or  variable. 

A  constant  vector  quantity  remains  unchanged  not  only  in  mag- 
nitude but  in  direction.  A  variable  vector  quantity  changes  either 
in  magnitude  or  in  direction,  or  in  both. 

If  a  point  A  is  fixed,  while  another  point  B  moves  in  any  man- 
ner, the  vector  AB  is  variable.  If  B  moves  always  along  the  straight 
line  which  at  a  certain  instant  passes  through  the  two  points,  the 
vector  varies  in  magnitude  only.  If  B  describes  a  circle  with 
center  at  A,  the  vector  varies  in  direction  but  not  in  magnitude. 
If  B  describes  any  other  path,  the  vector  varies  both  in  magnitude 
and  in  direction. 

22.  Increment  of  Variable  Vector. —  If  a  variable  vector  has 
different  values  at  the  beginning  and  end  of  any  interval  of  time,  its 
increment  during  that  interval  is  the  vector 

which,  added  to  the  initial  value,  will  pro- 
duce the  final  value.  The  increment  may 
be  determined,  when  the  initial  and  final 
values  of  the  vector  are  known,  by  the  fol- 
lowing construction: 

From  any  point  0  (Fig.  6),  lay  oft"  OA 
and  OB,  representing  the  initial  and  final  FlG*  6- 

values  of  the  variable  vector  ;    then    AB 

represents  the  required  increment.     For,  there  may  be  written  the 
vector  equation 

OA  +  AB=OB, 

showing  that  AB  satisfies  the  definition  of  increment  just  given. 

The  increment  is  evidently  found  by  subtracting  the  initial  value 
of  the  variable  vector  from  the  final  value.     For 

OB—OA=AB. 


12 


THEORETICAL    MECHANICS. 


In  general,  if  px  denotes  the  initial  value  of  a  vector  and  p,z  its  final 
value,  the  vector  increment  is  equal  to/2  — plt  the  minus  sign  de- 
noting vector  subtraction  (Art.  19). 

23.  Resolution  of  a  Vector. — To  resolve  a  vector  is  to  express 
it  as  the  sum  of  any  number  of  vectors.  If  AB  represents  the 
given  vector,  it  can  be  resolved  by  constructing  a  closed  polygon  of 
which  AB  is  a  side  ;  thus, 

AB=AL  +  LM+  MN  +  NBy 

L,  M,  and  N  being  any  three  points  whatever. 

Any  vectors  whose  sum  is  equal  to  a  given  vector  are  called 
components  of  that  vector. 

The  discussion  which  follows  will  be  limited  mainly  to  coplanar 
vectors  ;  that  is,  to  vectors  lying  in,  or  parallel  to,  the  same  plane. 
The  foregoing  principles  are,  however,  true  without  this  restriction. 

24.  Resolution  into  Two  Vectors  Having  Given  Directions. — 
A  given  vector  can  be  expressed  as  the  sum  of  two  vectors  parallel 

to  any  two  lines  coplanar  with  it.  For,  let 
AB  (Fig.  7)  represent  the  given  vector, 
and  MNy  PQ  the  given  lines.  From  A 
draw  a  line  parallel  to  MN>  and  from  B  a 
line  parallel  to  PQ.  Let  C  be  the  point  of 
intersection  of  these  lines  ;  then  A  C -\-  CB 
=  ABy  and  AC  and  CB  are  the  required 
components. 

An  equally  correct  construction  is  to 
draw  A  C  parallel  to  PQ,  and  CB  parallel 
to  MN.  The  result  is  the  same  as  before, 
since  A  C  and  CB  are  equal  vectors,  as 
are  also  AC  and  CB. 

Evidently  only  one  pair  of  components 
can  be  found  which  are  equivalent  to  a 
vector  and    have    assigned    directions. 

Resolved  Part  of  a  Vector.—  If  A  C 

and   CB  (Fig.  8)  are  at  right  angles  to  each 
other,  each  is  called  the  resolved  part  of  AB. 
If  AB  represents  any  vector  quantity,  and 
6  is  the  angle  between  AB  and  a  given  direc-  Fig.  8. 


Cr-~— 


given 
25 


INTRODUCTORY. 


13 


ma 


Fig. 


E'   N 


9- 


tion,  the  resolved  part  of  AB  in  that  direction  is  given  in  magnitude 
by  the  product 

(magnitude  AB)  X  (cos  6). 

The  resolved  part  of  a  vector  is  equal  to  the  algebraic  sum 
of  the  resolved  parts  of  its  components,  for  any  direction  of  reso- 
lution. Thus  (Fig.  9) 
AE  is  the  sum  of  the 
vectors  AB,  BC,  CD, 
DE.  The  resolved  parts 
of  these  components  par- 
allel to  the  line  MN  are 
A'B',  B'C,  CD', 
D'E' ;  while  the  re- 
solved part  of  AE  is 
A'E'.  Evidently  A'E' 
is  the  algebraic  sum  of 
A'B',  B'C,  CD',  D'E'. 

26.  Localized  Vector. —  If  a  vector  is  associated  with  a  definite 
point  or  line  in  space,  it  may  be  called  a  localized  vector. 

The  point  at  which  a  vector  is  localized  may  be  called  its  posi- 
tion; and  the  line  in  which  it  is  localized  its  position-line. 

27.  Moment  of  Localized  Vector.—  The  moment  of  a  localized 
vector  about  a  point  is  the  product  of  the  magnitude  of  the  vector 
into  the  perpendicular  distance  of  its  position-line  from  the  given 
point. 

The  origin  of  moments  is  the  point  about  which  the  moment 
is  taken.      The  arm  of  the  vector  is   the  perpendicular  distance 
of  its  position-line  from  the  origin. 

The  plane  of  the  moment  is  the  plane 
containing  the  origin  of  moments  and  the 
position-line  of  the  vector. 

In  comparing  moments  having  the  same 
plane  (or  parallel  planes)  there  are  two  cases 
which  may  be  distinguished  by  signs  plus 
and  minus.  These  cases  are  illustrated  in 
Fig.  10,  in  which  O  is  the  origin  of  mo- 
ments and  MN,  PQ  denote  the  directions  and  position-lines  of  two 
localized  vectors.     The  relations  of  MN  and  PQ  to  the  origin  O 


14  THEORETICAL    MECHANICS. 

suggest  rotations  about  O  in  opposite  directions.  One  of  these  di- 
rections being  taken  as  plus,  the  other  will  be  called  minus.  Which 
shall  be  called  plus  may  be  decided  arbitrarily.  For  uniformity,  it 
will  usually  be  assumed  that  the  positive  direction  is  counter-clock- 
wise ;  that  is,  opposite  to  the  direction  of  rotation  of  the  hands  of  a 
watch  placed  face  upward  in  the  plane  of  the  paper. 

28.  Moment  Represented  by  Area  of  Triangle. —  If  a  triangle 
be  constructed  with  its  vertex  at  the  origin  of  moments  and  its  base 
in  the  position-line  of  a  localized  vector,  the  length  of  the  base  being 
numerically  equal  to  the  magnitude  of  the  vector,  the  area  of  the 
triangle  is  numerically  equal  to  half  the  moment  of  the  vector.  This 
follows  immediately  from  the  definition  of  moment. 

29.  Moment  of  Sum  of  Two  Vectors. —  If/,  q  and  r  represent 
three  localized  vectors  such  that/  +  q  =  r>  and  if  their  position-lines 
meet  in  a  point,  the  algebraic  sum  of  the  moments  of  /  and  q  with 
respect  to  any  origin  in  their  plane  is  equal  to  the  moment  of  r  with 
respect  to  that  origin. 

Let  the  three  vectors  p,  q  and  r  be  represented  in  magnitude 
and  direction  by  OA,  OB  and   OC  (Fig.  11)  respectively,  so   that 


Fig.  11. 

OC  is  the  diagonal  of  the  parallelogram  of  which  OA  and  OB  are 
sides.  Let  O  be  the  point  of  intersection  of  the  three  position-lines, 
and  M  the  origin  of  moments.      Draw  MO.     Then 

moment  of  OA  =  2  (area  of  triangle  MOA); 
"  OB  =  2  (  "  "  "  MOB); 
"  OC=  2  (   "     "        "      MOC). 


INTRODUCTORY.  1 5 

Draw  AA\  BB ',    CC ',  each  perpendicular  to  MO.     The  double 
areas  of  the  three  triangles  are  respectively 

MO  X  AA\  MO  X  BB',  MO  X  CC 

Now,  ;         CC  =  AA'  +  BB'\        .         .        .     (i) 

for,  drawing  AC"  parallel  to  OM,  it  is  seen  that  A  A'  =  C"  C  and 
BB'=CC".     Hence 

area  MO  A  +  area  MOB  =  area  MOC.     .         .     (2) 

This  proves  the   proposition  for  the   case  in  which  the   moments   of 
the  two  given  vectors  have  the  same  sign. 

The  case  in  which  the  origin  is  so  taken  that  the  moments  have 
opposite  signs  is  represented  in  Fig.  1 2.     The  lettering  is  the  same 


as  in  Fig.  1 1 ,  and  the  reasoning  to  be  employed  is  exactly  similar. 
The  algebraic  sum  of  the  moments  is  now  the  numerical  difference. 
Thus,  from  the  geometry  of  the  figure,  we  have 

CC  =  AA'  -BB'        .        .        .     (Y) 

instead  of  equation  (1),  and  therefore 

area  MO  A  —  area  MOB  =  area  MOC,  .     (2' ) 

instead  of  equation  (2). 

30.  Vector  Quantities  in  Mechanics. —  Some  of  the  important 
quantities  dealt  with  in  the  science  of  Mechanics  are  vector  quantities. 
The  foregoing  brief  discussion  is  given  as  a  useful  introduction  to 
the  development  of  the  elementary  principles  of  Mechanics. 


PART  I.     STATICS. 


CHAPTER    II. 

FORCE    AND    STRESS. 

§  i.    Conception  of  Force. 

31.  Motion  of  a  Body  if  Uninfluenced  by  Other  Bodies. —  A 
body  remains  at  rest,  or  moves  in  a  straight  line  at  a  uniform  rate, 
except  in  so  far  as  its  motion  is  influenced  by  other  bodies.  (  New- 
ton's  first  law  of  motion.     See  Art.  259.) 

Since  no  body  can  be  wholly  free  from  the  influence  of  other 
bodies,  it  is  impossible  to  verify  this  law  directly.  The  indirect 
evidence  for  its  truth  is,  however,  very  strong.  When  a  body  is 
observed  to  depart  from  a  condition  of  uniform  rectilinear  motion,  it 
is  always  found  reasonable  to  attribute  this  departure  to  the  influence 
of  other  bodies.  Thus,  a  stone  thrown  horizontal^  into  the  air 
departs  from  the  condition  of  uniform  motion  in  a  straight  line  in  two 
ways:  its  path  curves  downward,  and  its  rate  of  hiotioVi  changes. 
The  first  effect  we  attribute  to  the  influence  of  the  earth  (which  we 
say  attracts  the  body),  while  the  second  we  attribute  partly  to  the 
"attraction"  of  the  earth  and  partly  to  the  influence  of  the  air 
(which  we  say  resists  the  motion  of  the  body). 

32.  Force  Defined. —  When  one  body  influences  the  motion  of 
another  body,  the  first  is  said  to  exert  a  force  upon  the  second. 
Force  may  therefore  be  defined  as  follows  : 

A  force  is  an  action  exerted  by  one  body  upon  another,  tending 
to  change  the  state  of  motion  of  the  body  acted  upon. 

When  an  external  body  exerts  a  force  upon  any  portion  of  the 
human  body,  the  action  (if  sufficiently  intense)  is  recognized  by  the 
senses.  In  such  a  case  the  name  pusk  or  pull 'is  given  to  the  force. 
It  is  therefore  natural  to  conceive  of  every  force  as  a  push  or  a  pull. 
This  conception  will  be  found  useful  in  deciding  what  may  properly 
be  called  forces. 


FORCE   AND    STRESS.  17 

33.  Magnitude  and  Direction  of  a  Force — When  the  action  of 
forces  is  recognized  by  the  senses,  we  form  the  idea  that  forces 
possess  different  magnitudes.  Thus,  if  two  pressures  are  successively- 
applied  to  the  same  part  of  the  body,  one  may  be  recognized  as 
much  greater  than  the  other.  We  thus  come  to  think  of  a  force  as 
possessing  a  definite  magnitude  ;  although  we  are  not  able  to  compare 
the  magnitudes  of  forces  with  precision  by  means  of  their  immediate 
effects  upon  the  senses. 

Experience  also  gives  the  idea  that  a  force  possesses  a  definite 
direction  ;  but  we  are  not  able,  from  the  immediate  evidence  of  the 
senses,  to  compare  the  directions  of  forces  with  accuracy. 

This  conception  of  a  force  as  possessing  definite  magnitude  and 
direction  is  verified  and  made  definite  when  the  effects  of  forces  upon 
the  motions  of  bodies  are  studied. 

34.  Force  a  Vector  Quantity.—  Since  a  force  possesses  both  a 
definite  magnitude  and  a  definite  direction,  it  is  a  vector  quantity 
(Art.  16). 

Like  other  vector  quantities,  forces  may  be  represented  by 
directed  right  lines  or  vectors.  To  accomplish  this,  some  length 
must  be  chosen  to  represent  the  unit  force  ;  then  any  given  force 
will  be  represented  by  a  vector  whose  length  is  the  same  multiple  of 
the  chosen  length  that  the  magnitude  of  the  force  is  of  the  unit  force. 

35.  Action  and  Reaction.—  If  a  body  A  exerts  a  force  upon  a 
second  body  B,  the  body  B  at  the  same  time  exerts  upon  A  a  force 
of  equal  magnitude  in  the  opposite  direction.  One  of  the  forces  of 
such  a  pair  is  often  called  an  action  and  the  other  a  reaction  ;  and 
the  principle  may  be  stated  in  the  words  *■ '  to  every  action  there 
corresponds  an  equal  and  contrary  reaction."  (Newton's  third  law 
of  motion.     See  Art.  259.) 

36.  Stress. —  A  stress  consists  of  two  equal  and  opposite  forces 
constituting  the  action  and  reaction  between  two  bodies  or  portions 
of  matter. 

Illustrations. — Any  two  bodies  attract  each  other  in  accordance 
with  the  law  of  gravitation  ;  the  two  forces  they  exert  upon  each 
other  constitute  a  stress.  Such  a  stress  acts  between  the  earth  and 
the  moon,  or  between  any  other  two  heavenly  bodies. 

Two  electrified  bodies  attract  (or  repel )  each  other  with  equal  and 
opposite  forces,  constituting  a  stress. 


1 8  THEORETICAL   MECHANICS. 

Two  bodies  in  contact  exert  upon  each  other  equal  and  opposite 
forces  at  the  surface  of  touching;  these  forces  constitute  a  stress. 

37.  Place  of  Application  of  a  Force.—  A  force  applied  to  a  body 
may  act  either  throughout  a  certain  volume,  or  upon  every  part  of  a 
certain  surface.  Thus,  the  attraction  of  the  earth  upon  a  body  ac- 
cording to  the  law  of  gravitation  acts  upon  every  part  of  the  mass, 
and  therefore  its  place  of  application  is  the  whole  space  occupied  by 
the  body;  while  the  pressure  between  two  bodies  which  touch  each 
other  is  distributed  over  their  surface  of  contact.  In  both  cases  the 
forces  are  distributed. 

A  concentrated  force  is  a  force  of  finite  magnitude  applied  at  a 
point.  Although  this  definition  is  ideal,  since  all  forces  are  in  real- 
ity distributed  in  their  action,  either  throughout  a  volume  or  over  an 
area,  the  conception  of  a  concentrated  force  is  a  useful  one.  In 
many  cases  a  relatively  large  force  is  applied  throughout  a  small 
volume  or  over  a  small  area,  and  the  conception  of  a  concentrated 
force  is  approximately  realized. 

By  the  study  of  the  effects  of  forces  upon  the  motions  of  the 
bodies  to  which  they  are  applied,  the  conception  is  reached  that  a 
distributed  force  is  ' '  equivalent  to  "  a  concentrated  force.  It  is  with 
the  laws  of  equivalence  of  systems  of  concentrated  forces  that  the 
science  of  Statics  mainly  deals;  a  distributed  force  being  regarded 
as  the  limiting  case  of  a  system  of  concentrated  forces  whose  number 
becomes  very  great  while  their  individual  magnitudes  become  very 
small. 

A  force  is  thus  regarded  as  having  a  definite  point  of  application 
and  a  definite  line  of  action.  It  is  thus  a  localized  vector  quantity 
(Art.  26). 

§  2.   Classes  of  Forces. 

38.  Conditions  Under  Which  Forces  Act.— The  conditions 
under  which  bodies  exert  forces  upon  one  another  are  known  only 
from  observation  and  experiment.  It  will  be  useful  to  enumerate  a 
few  cases  of  forces  whose  laws  have  been  the  subject  of  scientific 
investigation. 

39.  Examples  of  Forces. —  (1)  Gravitation. — Any  two  bodies 
exert  upon  each  other  attractive  forces  in  accordance  with  the  follow- 
ing law,  called  Newton' s  law  of  gravitation  : 


FORCE   AND    STRESS.  1 9 

Every  particle  of  matter  attracts  every  other  particle  with  a  force 
which  acts  in  the  line  joining  the  two  particles,  and  whose  magnitude 
is  proportional  directly  to  the  product  of  their  masses  and  inversely 
to  the  square  of  the  distance  between  them. 

(2)  Electrical  forces. — If  two  bodies  are  charged  with  electricity, 
each  exerts  a  force  upon  the  other.  The  forces  are  repulsive  or 
attractive,  according  as  the  charges  are  ' '  like  "  or  ' '  unlike ' '  in  kind. 

(3)  Magnetic  forces. — Two  magnets  exert  upon  each  other  forces, 
both  attractive  and  repulsive  ;  like  poles  repelling  and  unlike  attract- 
ing each  other. 

(4)  Electromagnetic  forces. — Forces  are  exerted  upon  each  other 
by  a  magnet  and  a  wire  carrying  an  electric  current;  also  by  two 
wires  carrying  electric  currents. 

(5)  Molecular  and  atomic  forces. — The  "atomic  theory"  of  the 
constitution  of  matter  assumes  that  a  body  of  definite  composition  is 
made  up  of  ultimate  particles  called  molecules,  each  of  which  possesses 
the  same  physical  properties  as  the  whole  body;  while  each  molecule 
is  made  up  of  lesser  parts  called  atoms,  which  do  not  separately 
possess  the  physical  properties  of  the  body.  The  theory  assumes 
that  forces  (besides  the  universal  force  of  gravitation)  act  between  the 
ultimate  particles. 

40.  Action  at  a  Distance. —  Forces  are  sometimes  classified  as 
"actions  at  a  distance"  and  "actions  by  contact."  Under  the 
former  would  be  classed  gravitational,  electrical,  and  magnetic  forces, 
since  they  apparently  are  not  exerted  by  means  of  any  material 
connection  between  the  bodies  concerned.  Modern  researches  have 
rendered  it  highly  probable  that  electrical  and  magnetic  forces  are 
in  reality  actions  transmitted  through  a  ' '  medium  ' '  which  fills  all 
space.  Efforts  have  also  been  made  to  show  that  the  force  of  gravi- 
tation may  be  explained  in  a  similar  manner.  Whether  this  view  be 
correct  or  not,  the  distinction  between  actions  at  a  distance  and  actions 
by  contact  is  practically  useful. 

41.  Passive  Resistances. —  The  forces  which  bodies  exert  upon 
one  another  are  often  divided  into  two  classes,  called  respectively 
active  forces  and  passive  resistances.  By  an  active  force  is  meant 
one  which  acts  independently  of  the  state  of  motion  of  the  body,  and 
also  independently  of  any  other  forces  which  may  be  applied  to  it  ; 
while  a  passive  resistance  comes  into  action  only  to  prevent  or  to 
resist  certain  motions  of  the  body. 


20  THEORETICAL    MECHANICS. 

Thus,  if  a  body  rests  upon  a  horizontal  table,  it  is  acted  upon  by 
the  attraction  of  the  earth,  which  is  an  active  force,  tending  to  give 
it  motion  downward,  and  having  the  same  magnitude  and  direction 
whatever  other  forces  may  be  applied  to  the  body  ;  and  also  by  a 
passive  resistance,  exerted  upward  by  the  table  to  resist  the  tendency 
of  the  body  to  move  downward.  This  latter  force  acts  only  because 
the  body  has  a  tendency  to  move  downward  through  the  table.  If, 
by  means  of  the  hand,  there  be  applied  to  the  body  a  supporting 
force  of  less  magnitude  than  the  downward  pull  of  the  earth,  the 
resisting  force  exerted  by  the  table  is  diminished,  its  amount  being 
always  just  sufficient  to  neutralize  the  effect  of  the  other  forces.  If 
the  hand  supports  the  body  with  an  upward  force  equal  to  the 
downward  attraction  of  the  earth,  the  pressure  of  the  table  upon  it 
becomes  zero.  If  the  upward  force  applied  by  the  hand  becomes 
still  greater,  the  table  exerts  no  force  to  resist  the  tendency  of  the 
body  to  rise.  This  illustrates  the  fact  that  a  passive  resistance  can 
have  any  magnitude  within  certain  limits,  but  that  its  value  cannot 
go  outside  these  limits. 

If  it  were  possible  to  analyze  the  forces  exerted  by  one  body  upon 
another  at  their  surface  of  contact,  and  by  contiguous  portions  of  the 
same  body  upon  each  other,  in  such  a  way  as  to  determine  the 
actions  between  the  ultimate  particles  and  the  laws  governing  these 
molecular  or  atomic  forces,  it  is  probable  that  these  would  be  found 
to  have  the  same  essential  nature  as  the  forces  called  active.  But 
for  practical  purposes  the  distinction  between  active  forces  and 
passive  resistances  is  often  useful. 

In  the  following  Articles  are  considered  two  cases  of  passive 
resistances  which  are  of  importance  in  the  discussion  of  the  problems 
of  Statics. 

42.  Pressure  Between  Bodies  in  Contact. —  If  two  bodies  touch 
each  other,  each  exerts  a  force  upon  the  other  at  the  surface  of 
contact.  These  contact  forces  are  passive  resistances,  and  (within 
certain  limits)  will  take  any  values  necessary  to  resist  certain  relative 
motions  of  the  bodies.  But  the  magnitudes  and  directions  of  the 
forces  which  the  bodies  can  exert  upon  each  other  are  subject  to 
certain  limitations  depending  upon  the  nature  of  the  surfaces  and  the 
material  composing  the  bodies. 

Let  A  and  B  (Fig.  1 3)  represent  the  two  bodies.  If  for  any 
reason  A   has   a   tendency  to  slide  upon  B,  this  tendency  will  be 


FORCE    AND    STRESS.  21 

resisted,  and  may  be  wholly  neutralized.  The  rougher  the  surfaces 
of  contact,  the  greater  the  force  that  can  be  exerted  to  resist  the 
sliding.  If  the  surfaces  be  very  smooth,  the  possible  magnitude  of 
the  force  is  very  small.  Hence  we  are  led  to  the  fol- 
lowing definition  : 

A  perfectly  smooth  surface  is  one  which  can  offer 
no  resistance  to  the  sliding  of  a  body  upon  it. 

Although  this  definition  cannot  be  realized  in  the 
Fig.  13.         case  °f  anv  actual  body,  certain  surfaces  approach 

near  to  the  condition  of  perfect  smoothness. 
If  for  any  reason  A  has  a  tendency  to  move  toward  B  in  the 
direction  of  the  normal  to  their  surface  of  contact,  the  latter  body  is 
able  to  exert  a  resisting  force  in  the  direction  of  the  normal.  The 
magnitude  which  this  resisting  force  can  take  is  limited  only  by  the 
strength  of  the  material  of  which  B  is  composed.  Such  a  ' 4  normal 
resistance ' '  can  be  exerted  by  a  smooth  surface  as  well  as  by  a  rough 
one.  The  pressure  exerted  between  rough  surfaces  will  be  consid- 
ered later  under  the  head  of  friction. 

Smooth  hinge. — A  hinge  joint  is  often   used  to   connect  two 
bodies,  in  such  a  way  as  to  leave  them  free  to  move  in  a  certain 
manner  relatively  to  one  another.     Such  a  joint  consists  of  a  cylin- 
drical pin  which  forms  a  part 
of  one   body,   or   is  rigidly 
connected  with  it,  and  a  cy- 
lindrical hole  formed  in  the 
other  body,  into  which  the 
pin  is  inserted.     Such  a  con- 
nection permits  the  bodies  to 
turn  relatively  to  each  other  ■FlG-  I4' 

about  the  axis  of  the  pin,  but 

(if  the  pin  and  hole  are  of  equal  diameter)  prevents  any  other  rela- 
tive motion.  (See  Fig.  14.)  In  order  to  permit  free  motion  about 
the  axis,  the  pin  must  be  slightly  smaller  than  the  hole,  so  that  the 
two  bodies  are  in  contact  along  a  straight  element  common  to  the 
two  cylindrical  surfaces.  If  these  surfaces  are  smooth,  the  pressure 
between  them  has  the  direction  of  their  common  normal.  In  the 
problems  of  Statics,  a  smooth  hinge  is  often  introduced  as  a  means  of 
connecting  bodies.  In  such  a  case  the  pressure  between  the  bodies 
is  to  be  taken  as  acting  in  a  line  through*  the  center  of  the  hinge. 


22  THEORETICAL    MECHANICS. 

43.  Tension  in  a  Flexible  Cord.—  A  flexible  cord  is  one  which 
may  be  bent.  A  perfectly  flexible  cord  may  be  bent  without  resist- 
ance. No  actual  cord  possesses  perfect  flexibility,  but  the  resistance 
to  bending  may  be  very  slight. 

Let  AB  (Fig.  15)  represent  a  portion  of  a  perfectly  flexible 
cord,  and  let  two  equal  and  opposite  forces,  Pand  P' ,  be  applied  to 
it  at  A  and  B.     Let  C  be  any  point  between  A  and  B.     The  force  P 

tends  to  cause  AC  to 

P^        i\ww\!\  \  \\m     _  yiP'    move  to  the  left  ;  this 

A.  C  B  is  prevented  by  a  force 

Fig.  15.  equal  and  opposite  to 

P  exerted  at  C  by  CB 
upon  AC.  Similarly,  P'  tends  to  cause  CB  to  move  to  the  right ; 
and  this  is  prevented  by  a  force  equal  and  opposite  to  P'  exerted  at 
C  by  A  C  upon  CB.  That  is,  the  two  parts  of  the  cord  exert  upon 
each  other  at  C  equal  and  opposite  forces,  constituting  a  stress  (Art. 
36).  The  same  is  true  at  any  other  point  between  A  and  B.  Such 
a  stress  (resisting  a  tendency  of  the  two  portions  of  the  cord  to  sep- 
arate in  the  direction  of  their  length)  is  called  a  tension. 

We  have  here  illustrated  a  special  case  of  the  principle  that  if  a 
flexible  cord  is  pulled  tight  between  two  points,  any  two  adjacent 
portions  exert  upon  each  other  forces  par- 
allel to  the  direction  of  the  cord. 

This  subject  will  be  discussed  more 
fully  in  a  later  chapter.  It  is  introduced 
at  this  point  because  in  the  problems  to 
be  discussed  in  the  following  chapters  a 
flexible  cord  is  often  used  as  a  means  of 
applying  a  force  to  a  body.     Thus,  if  a 

cord  is  attached  to  a  body  X  (Fig.  16)  at  a  point  By  a  force  of  any 
magnitude  applied  to  the  cord  at  another  point  A,  in  the  direction 
BA}  will  cause  the  cord  to  exert  an  equal  force  upon  the  body  at  B. 

§  3.  Numerical  Representation  of  Forces  and  of  Masses —  Grav- 
itation System. 

44.  Relation  Between  Units  of  Force  and  of  Mass. —  Although 
force  and  mass  are  two  distinct  kinds  of  quantity,  so  that  it  is  pos- 
sible to  choose  the  unit  in  which  each  is  expressed  independently  of 


FORCE    AND    STRESS.  23 

the  other,  yet  it  is  convenient  to  choose  these  units  so  that  one  de- 
pends upon  the  other.  The  method  of  choosing  units  which  will 
now  be  described  makes  the  unit  mass  arbitrary,  while  the  unit  of 
force  depends  upon  that  of  mass.  This  is  the  gravitation  system  of 
units.     In  a  later  chapter  another  system  will  be  described. 

45.  Unit  Mass. —  The  unit  mass  is  taken  as  the  quantity  of  mat- 
ter in  a  certain  piece  of  platinum  arbitrarily  chosen  and  established 
as  the  standard  by  act  of  the  British  Parliament.  This  unit  mass  is 
called  one  pound. 

46.  Weight. — The  weight  of  a  body  is  the  force  with  which  the 
earth  attracts  it,  in  accordance  with  the  law  of  gravitation.  From 
this  law  (Art.  39)  it  follows  that  the  weights  of  bodies  in  the  same 
locality  at  the  surface  of  the  earth  are  proportional  to  their  masses. 
But  if  the  bodies  are  in  different  positions  on  the  earth,  or  at  differ- 
ent elevations  above  the  surface,  their  weights  may  not  be  propor- 
tional to  their  masses. 

In  problems  relating  to  bodies  near  the  earth,  the  weight  of  a 
body  has  often  to  be  included  among  the  forces  applied  to  it.  By 
the  law  of  gravitation  every  particle  of  the  earth  attracts  every  par- 
ticle of  the  body.  These  several  forces  may  for  many  purposes  be 
regarded  as  equivalent  to  a  single  force  applied  at  a  certain  point 
called  the  center  of  gravity  of  the  body.  The  truth  of  this  state- 
ment is  for  the  present  assumed  without  proof.     (See  Chapter  IX.) 

47.  Unit  Force. —  The  gravitation  unit  of  force is  a  force  equal 
to  the  weight  of  a  unit  mass  at  the  earth's  surface. 

A  pound  force  is  a  force  equal  to  the  weight  of  a  pound  mass  at 
the  earth's  surface. 

The  pound  force  as  thus  defined  has  not  the  same  value  for  all 
positions  on  the  surface  of  the  earth,  since  the  weight  of  any  given 
body  varies  if  it  be  taken  to  different  localities.  The  variations  are, 
however,  comparatively  small,  and  for  many  purposes  unimportant. 
The  pound  force  may  be  made  wholly  definite  by  specifying  a  certain 
place  at  which  its  value  is  to  be  determined. 

It  is  unfortunate  that  the  word  pound  is  used  in  two  senses,  ap- 
plying both  to  a  unit  force  and  to  a  unit  mass.  The  usage  is,  how- 
ever, so  common  that  it  is  important  for  the  student  to  become  fa- 
miliar with  it.     The  double  meaning  arises  from  the  fact  that  the 


24  THEORETICAL    MECHANICS. 

most  convenient  as  well  as  the  most  accurate  method  of  comparing 
the  masses  of  bodies  is  by  determining  the  ratio  of  their  weights. 

48.  Comparison  of  Forces. —  The  magnitude  of  a  force  in  terms 
of  the  gravitation  unit,  or  pound  force,  may  be  found  by  determin- 
ing how  many  pounds  mass  it  will  support  against  the  attraction  of 
the  earth.  This  is  the  method  employed,  directly  or  indirectly,  in 
many  machines  for  testing  the  strength  of  materials. 

49.  Measurement  of  Masses. —  The  mass  of  a  body  may  be 
determined  by  comparing  its  weight  with  that  of  a  body  or  bodies 
of  known  mass.  This  is  the  method  adopted  in  ' '  weighing ' '  a  body 
on  an  ordinary  balance  consisting  of  a  lever  or  a  system  of  levers. 
The  weight  of  a  body  as  thus  determined  is  a  correct  indication  of 
its  mass,  provided  the  masses  of  the  standard  bodies  or  ' '  weights ' ' 
are  accurately  known;  for  if  the  same  experiment  be  conducted  in 
two  different  localities,  the  weights  of  the  standard  bodies  will  change 
in  the  same  ratio  as  the  weights  of  the  bodies  balanced  against  them. 

If  the  weight  of  a  body  is  determined  by  a  spring  balance,  the 
result  may  be  different  if  the  weighing  is  done  in  different  localities; 
for  in  this  case  the  process  of  weighing  consists,  not  in  comparing 
the  weight  of  one  body  with  that  of  another,  but  in  determining  di- 
rectly the  pull  exerted  by  the  earth  *  upon  the  body  by  ascertaining 
how  much  this  pull  will  stretch  a  spring.  If  the  weight  of  the  body 
changes,  therefore,  as  it  may  if  its  location  is  changed,  this  change 
will  be  indicated  by  a  corresponding  variation  in  the  amount  of 
stretching  of  the  spring. 

50.  Metric  System. —  In  the  gravitation  system  of  estimating 
forces  and  masses,  the  unit  mass  is  wholly  arbitrary.  Besides  the 
units  above  described,  which  owe  their  establishment  to  British  law 
and  custom,  there  is  another  important  set,  based  upon  the  French 
gram  or  the  kilogram.  A  gram  is  very  nearly  equal  to  the  mass  of 
a  cubic  centimeter  of  pure  water  at  the  temperature  of  maximum 
density.  This  relation  is  not  exact,  however,  and  the  gram,  like  the 
pound,  must  be  regarded  as  an  arbitrary  unit,  whose  value  is  to  be 

*  What  is  actually  measured  is  the  ' '  apparent ' '  pull  of  the  earth  upon  the 
body,  which  differs  from  the  true  attractive  force  for  reasons  which  cannot  be 
here  discussed,  the  chief  of  them  being  the  diurnal  rotation  of  the  earth.  See 
Art.  311. 


FORCE    AND    STRESS.  25 

determined  by  reference  to  a  certain  standard  body.  For  ordinary 
purposes  the  kilogram,  equal  to  1,000  grams,  is  often  more  con- 
venient than  the  gram. 

The  gravitation  unit  of  force,  in  the  French  or  metric  system, 
may  be  taken  as  the  weight  of  a  body  of  one  kilogram  mass.  For 
brevity  this  unit  may  be  called  a  force  of  one  kilogram.  To  make 
the  unit  definite,  the  position  must  be  specified,  since  the  weight  of 
the  same  body  is  different  in  different  places. 

The  following  is  the  relation  between  the  pound  and  the  kilo- 
gram : 

1  kilogram  =  2.2046    pounds. 
1  pound       =  0.45359  kilogram. 

The  gravitation  system  of  units  of  force  and  mass,  although  the 
most  convenient  in  many  practical  applications,  is  not  the  best  for  the 
purposes  of  pure  science.  The  discussion  of  other  systems  must, 
however,  be  deferred. 


§  4.   Definitions. 

51.  Concurrent  and  Non-concurrent  Forces. —  Forces  acting 
upon  a  body  are  concurrent  when  they  have  the  same  point  of  appli- 
cation.    When  applied  at  different  points  they  are  non- concurrent. 

52.  Coplanar  Forces  are  those  whose  lines  of  action  lie  in  the 
same  plane.     The  following  pages  treat  mainly  of  coplanar  systems. 

53.  A  Couple  is  a  system  consisting  of  two  forces,  equal  in  mag- 
nitude, opposite  in  direction,  and  having  different  lines  of  action. 
The  perpendicular  distance  between  the  two  lines  of  action  is  called 
the  arm  of  the  couple. 

54.  Equivalent  Systems  of  Forces. —  Two  systems  of  forces  are 
equivalent  if  either  may  be  substituted  for  the  other  without  change 
of  effect. 

From  this  definition  it  follows  that  two  systems,  each  of  which  is 
equivalent  to  a  third  system,  are  equivalent  to  each  other. 

55.  Resultant. —  A  single  force  that  is  equivalent  to  a  given  sys- 
tem of  forces  is  called  the  resultant  of  that  system.  It  will  be  shown 
subsequently  that  a  system  of  forces  may  not  be  equivalent  to  any 


26  THEORETICAL    MECHANICS. 

single  force.     When  such  is  the  case,  the  simplest  system  equivalent 
to  the  given  system  may  be  called  its  resultant. 

Any  forces  having  a  given  force  for  their  resultant  are  called 
components  of  that  force. 

56.  Composition  and  Resolution  of  Forces. —  Having  given 
any  system  of  forces,  the  process  of  finding  an  equivalent  system  is 
called  the  composition  of  forces  if  the  system  determined  is  simpler 
than  the  given  system ;  if  the  reverse  is  the  case,  the  process  is  called 
the  resolution  of  forces. 

The  process  of  finding  the  resultant  of  any  given  forces  is  the 
most  important  case  of  composition ;  while  the  process  of  finding  two 
or  more  forces,  which  together  are  equivalent  to  a  single  given  force, 
is  the  most  common  case  of  resolution. 

57.  Equilibrium. —  A  system  of  forces  applied  to  a  body  is  in 
equilibrium  if  the  state  of  motion  of  the  body  is  not  changed  by 
their  action. 

The  term  equilibrium  is  also  applied  to  the  body  upon  which  the 
forces  act  ;  a  body  is  said  to  be  in  equilibrium  if  its  state  of  motion 
is  unchanged  by  the  action  of  all  applied  forces. 


CHAPTER   III. 

CONCURRENT    FORCES. 

§  i.    Composition  and  Resolution  of  Concurrent  Forces. 

58.  Forces  Having  the  Same  Line  of  Action.—  The  resultant  of 
two  concurrent  forces  acting  in  the  same  direction  is  a  force  in  that 
direction  equal  to  their  sum. 

The  resultant  of  two  concurrent  forces  acting  in  opposite  directions 
is  a  force  whose  magnitude  is  equal  to  the  difference  between  the 
magnitudes  of  the  given  forces,  and  whose  direction  is  that  of  the 
greater. 

No  proof  need  be  given  of  the  truth  of  these  statements.  They 
must  be  regarded  as  following  immediately  from  our  conception  of 
force  and  from  our  experience  regarding  the  effects  of  forces.  They 
are  special  cases  of  the  general  principle  of  the  parallelogram  of 
forces  (Art.  59). 

These  principles  may  be  generalized  in  the  following  manner  : 

( 1 )  The  resultant  of  any  number  ot  concurrent  forces  having  the 
same  direction  is  a  force  in  that  direction  whose  magnitude  is  equal 
to  the  sum  of  the  magnitudes  of  the  given  forces. 

(2)  The  resultant  of  any  concurrent  forces  having  the  same  line 
of  action  is  found  by  combining  those  having  one  direction  into  a 
single  force  and  those  having  the  opposite  direction  into  another 
force,  and  then  combining  the'se  partial  resultants  according  to  the 
rule  already  stated. 

If  signs  plus  and  minus  are  used  to  distinguish  the  two  opposite 
directions  along  the  common  line  of  action  of  the  forces,  the  foregoing 
principles  may  be  included  in  the  following  general  rule: 

The  resultant  of  any  concurrent  forces  having  the  same  line  of 
action  is  a  force  equal  to  their  algebraic  sum. 

59.  Resultant  of  Any  Two  Concurrent  Forces. —  If  any  two 

concurrent  forces  be  represented  in  magnitude  and  direction  by  lines 
drawn  from  the  same  point,  their  resultant  is  represented  in  magnitude 
and  direction  by  the  diagonal  (drawn  from  that  point)  of  the  parallel- 
ogram of  which  the  two  lines  are  adjacent  sides. 


28  THEORETICAL    MECHANICS. 

Thus  let  OA  and  OB  (Fig.  17)  represent  the  two  given  forces, 
and  OC  the  diagonal  of  the  parallelogram  of  which  OA  and  OB  are 

sides  ;  then  OC  represents  the  resultant. 
This  principle  is  known  as  the  par- 
allelogram of  forces.  Many  proofs  of 
the  proposition  have  been  offered,  but 
none  of  these  has  been  generally  accept- 
ed as  satisfactory.  Perhaps  the  most 
satisfactory  view  is  that  which  regards 
this  principle  as  one  of  the  fundamental 
laws  of  force,  depending  for  its  verifica- 
tion upon  experience.  The  matter  is  further  considered  in  Chapter 
XIV. 

60.  Triangle  of  Forces. —  Since  the  opposite  sides  of  a  parallel- 
ogram are  equal,  the  magnitude  and  direction  of  the  resultant  of 
two  forces  may  be  found  by  constructing  a  triangle  instead  of  a 
parallelogram.  Thus,  any  two  forces  being  given,  let  them  be 
represented  in  magnitude  and  direction  by  lines  OA  and  A  C  laid  off 
consecutively  (Fig.  17)  ;  then  OC  represents  the  magnitude  and 
direction  of  the  resultant.  For,  if  OB  be  drawn  equal  and  parallel 
to  AC,  OC  is  a  diagonal  of  the  parallelogram  of  which  OA  and 
OB  are  adjacent  sides.     The  lines  OB,  BC 

are  not  needed   for  the  determination  of 
OC.     Evidently,  the  principle  of  the  par- 
allelogram  of  forces   is  equivalent  to  the    . 
following  : 

If  A,  B,  C  are  three  points  so  chosen  Fig.  18. 

that  AB,  BC  represent,  in  magnitude  and 

direction,   two  concurrent  forces,  their  resultant  is  represented  in 
magnitude  and  direction  by  A  C. 

This  principle  is  known  as  the  triangle  of  forces. 

The  point  of  application  of  the  resultant  is  the  same  as  that  of 
the  components. 

61.  Computation  of  Resultant  of  Two  Concurrent  Forces. —  Let 
P  and  Q  represent  the  magnitudes  of  two  concurrent  forces,  and 
6  the  angle  between  their  lines  of  action.  [The  angle  is  to  be 
measured  between  the  positive  directions  of  the  lines  of  action.  Thus, 
if  XX'  and  YY'  (Fig.  19)  are  the  lines  of  action,  the  directions 


CONCURRENT    FORCES. 


29 


being  as  indicated  by  the  arrows,  0  is  the  angle  X'OY'.]  Laying 
offAB,  BC  to  represent  Pand  Q,  A  C  will  represent  the  resultant  R. 
By  elementary  Trigonometry,  the  magnitude  of  the  resultant  is  given 
by  the  formula 

R*  =  P*  +  Q>+  2pQ  cos  e. 

To  find  the  direction  of  the  resultant,  let  angle  CAB  =  a,  angle 
ACB  =  $\  then 

sin  a sin  fi sin  6 


or 


sin  a 


Q         P 

Q  P 

—  sin  6  ;     sin  8  —  —  sin  0. 
R  R 


Examples. 

1.  A  particle  is  acted  upon  by  forces  of  10  lbs.  and  20  lbs.  whose 
lines  of  action  include  an  angle  of  25°. 

Determine  the  magnitude  and  direc- 
tion of  a  single  force  which  would 
produce  the  same  effect. 

Ans.  A  force  of  29.4  lbs.  making 
an  angle  of  16°  42'  with  the  force  of 
10  lbs. 

2.  Two  men  pull  a  body  horizon- 
tally by  means  of  ropes.  One  exerts 
a  force  of  28  lbs.  directly  north,  the 
other  a  force  of  42  lbs.  directed  N. 
42°  E.  What  single  force  would  be 
equivalent  to  the  two  ? 

3.  Two  persons  lifting  a  body  ex- 
ert forces  of  44  lbs.  and  60  lbs.  in  di- 
rections inclined  28°  to  the  vertical  on  opposite  sides, 
force  would  produce  the  same  effect  ? 

Ans.  A  force  of  92. 1  lbs.  at  angle  320  40'  with  force  of  44  lbs. 


Fig. 


19. 


What  single 


62.  Resultant  of  Any  Number  of  Concurrent  Forces. —  The 

resultant  of  any  number  of  concurrent  forces  may  be  found  by  first 
finding  the  resultant  of  two  of  them,  then  combining  this  resultant 
with  a  third  force,  and  so  on. 

Thus,  if  AB  and  BC  (Fig.  20)  represent  two  of  the  forces  in 
magnitude  and  direction,  their  resultant  is  represented  by  A  C  Draw 
CD  to  represent  a  third  force;  the  resultant  of  AC  and  CD  is  AD, 
which  is  therefore  the  resultant  of  the  three  forces  represented  by 


30  THEORETICAL    MECHANICS. 

AB>  BC,  and  CD.     If  DE  represents  a  fourth  force,  the  resultant  of 
AD  and  DE  is  AE,  which  is  therefore  the  resultant  of  AB,  BC,  CD 

and  DE.     This  process  may  be  ex- 
Z?/^^^  tended    to    include   any    number   of 

/     ^^"^^^^       forces.     Evidently  the  lines  A  C  and 

Jr  '"/         A®    are   not    needed    m    tne   con" 

/  ^  +?"*   /  struction. 

/    „*""        ^    __     -p  The   above  process   may   be   de- 

a  ^y-- ~/~^r^^  scribed  as  follows  : 

"^^^^^^^  If  any  number  of  concurrent  forces 

D  are  represented  in  magnitude  and  di- 

Fig.  20.  rection   by  lines  forming  sides   of  a 

continuously  described  polygon,  the 

line  drawn  from  the  starting  point  to  close  the  polygon  represents 

the  resultant  in  magnitude  and  direction. 

This  principle  is  called  the  polygon  of  forces. 
The  sides  of  the  polygon  cannot,  of  course,  represent  the  lines  of 
action  of  the  forces,  since  these  all  intersect  in  the  point  at  which  the 
forces  are  applied. 

63.  Resultant  as  Vector  Sum. — The  process  of  finding  the  result- 
ant of  several  forces  by  constructing  the  polygon  is  evidently  the 
same  as  the  process  of  vector  addition  (Art.  18).     That  is, 

The  resultant  of  any  number  of  concurrent  forces  is  a  force  equal 
to  their  vector  sum,  its  point  of  application  being  the  same  as  that 
of  the  given  forces. 

64.  Vector  Diagrams  and  Space  Diagrams. — In  combining  forces 
by  geometrical  construction,  they  must  be  represented  in  magnitude 
and  direction,  and  also  in  line  of  action.  In  most  cases  it  is  desir- 
able, in  order  to  prevent  confusion,  to  draw  two  separate  diagrams, 
one  showing  the  lines  of  action  of  the  forces,  the  other  representing 
them  in  magnitude  and  direction  only.  These  two  classes  of  dia- 
grams may  be  called  space  diagrams  and  vector  diagrams  respec- 
tively. 

65.  Computation  of  Resultant  by  Force  Polygon. —  The  re- 
sultant of  any  coplanar  concurrent  forces  may  be  computed  by 
graphical  construction,  by  drawing  the  force  polygon  to  scale  and 
measuring  the  line  representing  the  resultant.  Or,  the  length  and 
direction  of  the  closing  side  of  the  force  polygon  may  be  computed 


CONCURRENT   FORCES.  3 1 

trigonometrically  from  the  force  polygon.     A  more  convenient  trig- 
onometric method  will  be  given  later. 

66.  Resolution  of  a  Force  Into  Any  Number  of  Compo- 
nents.—  A  force  may  be  resolved  into  any  number  of  components 
by  means  of  the  polygon  of  forces.  If  the  magnitude  and  direction 
of  the  given  force  be  represented  by  a 

line,  this  line  may  be  made  one  side  of  a 
polygon,  and  the  other  sides  will  repre- 
sent, in  magnitude  and  direction,  forces 
of  which  the  given  force  is  the  result- 
ant. Thus,  let  the  given  force  be  rep- 
resented by  the  vector  AB  (Fig.  21). 
Choose  any  number  of  points  (as  C,  D> 
E),  and  draw  lines  forming  the  poly- 
gon AC  DEB.  Then  AQ  CD,  DE} 
EB  represent,  in  magnitude  and  direc- 
tion, four  forces  whose  resultant  is  AB.  Fig.  21. 

A  force  may  obviously  be  replaced 
by  other  forces  in  an  infinite  number  of  ways.     Some  special  cases 
of  the  general  problem  of  resolution  will  now  be,  considered. 

67.  Resolution  of  a  Force  Into  Two  Components. —  This  is 
a  special  case  of  the  general  problem  of  resolution  into  any  number 
of  components.  If  a  force  be  represented  in  magnitude  and  direc- 
tion by  a  vector  AB,  and  if  C  be  any  point  whatever,  the  vectors 
A  C,  CB  represent  two  forces  which  together  are  equivalent  to  the 
given  force.  Since  an  infinite  number  of  triangles  may  be  drawn, 
each  having  any  given  line  as  one  side,  it  follows  that  a  force  may  be 
replaced  by  two  forces  in  an  infinite  number  of  ways. 

To  make  the  problem  determinate,  certain  conditions  must  be 
specified  which  the  components  are  to  satisfy.  The  following  cases 
furnish  determinate  problems.  Each  should  be  solved  both  by  geo- 
metrical construction  and  by  algebraic  computation. 

(1)  Let  a  force  be  resolved  into  two  components  whose  lines  of 
action  are  given. 

Solution,  (a)  Geometrical. —  Let  AB  (Fig.  22)  represent  the 
given  force  in  magnitude  and  direction,  and  let  the  two  components 
have  lines  of  action  parallel  to  XX  and  YY.  From  A  draw  a  line 
parallel  to  XX,  and  from  B  a  line  parallel  to  YY,  C  being  the 
point  of  intersection  of  these  two  lines;  then  AC  and  CB  represent 


32 


THEORETICAL    MECHANICS. 


the  required  components  in  magnitude  and  direction, 
tion  may  evidently  be  varied  by  drawing  from  A  a 
YY  and  from  B  a  line  parallel  to  XX, 
giving  A  C  and  C'B  as  the  two  com- 
ponents. The  result  is  the  same  as 
before,  since  AC  and  C'B  are  equal 
vectors,  as  are  also  A  C  and  CB. 

(£)  Algebraic. — From  the  given  data 
there  are  known  in  the  triangle  ACB 
the  side  AB  and  all  the  angles.  Hence, 
if  AB  =  R,  AC=Q,  CB=P,  ACB 
=  i8o°  —  0,  ABC=ay  BAC=frwe 
have  (as  in  Art.  6i) 


The  construc- 
line  parallel  to 


q  _  p 

sin  a       sin  /3 


A' 


sin  6 


from  which  P  and  Q  may  be  computed. 

(2)  Let  the  two  components  be  given 
in  magnitude  only. 

(3)  Let  the  magnitude  of  one  and  the  direction  of  the  other  be 
given. 

(4)  Let  one  component  be  wholly  given. 


Examples. 

1.  A  force  of  50  lbs.  is  equivalent  to  two  forces  whose  directions 
are  inclined  to  that  of  the  given  force  at  angles  of  1 20  and  86°  respec- 
tively; determine  their  magnitudes.     Ans.  50.6  lbs.  and  10.5  lbs. 

2.  A  force  of  83  lbs.  is  to  be  replaced  by  two  components,  one 
of  which  is  a  force  of  36  lbs.  at  right  angles  to  the  given  force.  De- 
termine the  magnitude  and  direction  of  the  other  component. 

3.  Resolve  a  force  of  200  lbs.  into  two  components,  of  130  lbs. 
and  98  lbs.  respectively.  Determine  the  directions  of  the  two  com- 
ponents. 

An s.  Angle  between  resultant  and  greater  component  =  24°  34'. 

4.  A  force  of  1 50  lbs.  is  to  be  resolved  into  two  components,  one 
acting  at  an  angle  of  20°  with  the  given  force,  the  other  having  a 
magnitude  of  80  lbs.      Determine  completely  the  two  components. 

5.  A  force  of  150  lbs.  is  to  be  resolved  into  two  components,  one 
acting  at  an  angle  of  20°  with  the  given  force.  What  is  the  least 
possible  magnitude  of  the  other?  Ans.  51.31  lbs. 


CONCURRENT    FORCES.  33 

68.  Resolved  Part  of  a  Force — Definition  .—If  a  force  is  equiv- 
alent to  two  components  at  right  angles  to  each  other,  each  is  called 
a  resolved  part  of  the  given  force. 

The  resolved  part  of  a  force  represented  by  AB  (Fig.  23)  in  the 
direction  of  a  line  MN  is  represented  in  magnitude  and  direction  by 
A'B',  the  orthographic  projection  of 
AB  upon  MN 

Proposition. —  The  resolved  part, 
in  a  given  direction,  of  the  resultant 
of  any  concurrent  forces,  is  equal  to 
the  algebraic  sum  of  the  resolved 
parts  of  the  components  in  that  di-  ft'     j\r 

rection.  Fig.  23. 

Since  the  resultant  is  the  vector 
sum   of  the  components,   this  proposition  is  a  special  case  of  that 
stated  in  Art.  25. 

Algebraic  expression  for  the  resolved  part. —  From  the  definition 
it  follows  that  the  resolved  part  of  a  force  of  magnitude  P  in  a  direc- 
tion making  an  angle  0  with  that  of  the  force  is  equal  to  P  cos  6. 
If  6  lies  between  90°  and  2700  its  cosine  is 
negative.  Hence  if  6  is  measured  from  the 
positive  direction  of  the  line  along  which  the 
resolution  is  made,  the  product  P  cos  6  gives 
the  resolved  part  with  proper  sign,  whatever 
the  value  of  the  angle. 

Let  X  and  Y  be  the  resolved  parts  of  a 
force  P  in  two  directions  at  right  angles  to 
each  other,  0  being  the  angle  between  P  and  JC,  and  7r/2  —  6  the 
angle  between  P  and  Y.     Then 

X  =  Pcos  6; 

Y  =  P  cos  (tt/2  —  6)  =  Psin  6; 

P*  =  X'1  +  F2; 

a       X       "     a        Y 
cos  v  =  — ;    sin  v  =  —  . 

P  P 

From  these  equations  we  may  compute  the  resolved  parts  when  the 
force  and  the  directions  of  resolution  are  given;  or  we  may  compute 
the  magnitude  and  direction  of  the  force  when  its  resolved  parts  in 
two  directions  at  right  angles  to  each  other  are  known. 


34  theoretical  mechanics. 

Examples. 

i.  Let  20  lbs.  and  40  lbs.  be  the  resolved  parts  of  a  force  in  two 
directions  at  right  angles  to  each  other.  Determine  the  magnitude 
and  direction  of  the  force. 

2.  Compute  the  resolved  part  of  a  force  of  235  lbs.  in  a  direction 
inclined  25°  to  that  of  the  force  ;  also  in  a  direction  inclined  160°  to 
that  of  the  force. 

69.  Algebraic  Computation  of  Resultant  of  Any  Concurrent 
Forces. —  From  the  principle  of  the  triangle  of  forces,  it  13  possible 
to  compute  the  magnitude  and  direction  of  the  resultant  of  any 
known  concurrent  forces.  Thus,  the  resultant  of  any  two  may  be 
computed  by  the  solution  of  a  triangle  ;  this  resultant  may  be  com- 
bined with  a  third  force  and  their  resultant  computed  in  a  similar 
manner;  and  the  process  may  be  continued  until  the  resultant  of  the 
whole  system  is  determined.  This  process  would,  however,  be  very 
laborious,  and  in  most  cases  the  following  method  is  to  be  preferred. 

Choose  a  pair  of  rectangular  axes  in  the  plane  of  the  forces,  and 
let  the  given  forces  be  Px ,  making  angles  ax ,  ft  with  the  axes  ;  P2 , 
making  angles  a2 ,  ft2  with  the  axes  ;  etc.  Replace  each  force  by  its 
resolved  parts  parallel  to  the  two  axes  ;  Px  being  replaced  by  com- 
ponents Px  cos  ax>  Px  cos  ft;  Pt  by  components  P2  cos  a.it  P2  cos  f&%\ 
etc.  The  given  forces  are  thus  replaced  by  two  systems  of  collinear 
forces.     The  resultants  of  these  two  systems  are  respectively 

Px  cos  ax  -\-  P.z  cos  a  ,  -f-      .      .      .      =  X, 

and  Px  cos  ft  +  P,  cos  £,  +     .     .     .     ==  Y. 

The  resultant  of  the  whole  system  is  equal  to  the  resultant  of  X  and 
Y.  Let  R  denote  this  resultant  and  a,  b  the  angles  it  makes  with  the 
x-  and  ^/-axes  respectively;  then 


r*  =  x*  +  y 


*     x         /,      Y 

cos  a  =  —  ;     cos  0  —  — . 
R  R 


Examples. 


1.  Find  the  resultant  of  the  following  concurrent  forces:  23  lbs. 
directed  N.  40°  E. ;  42  lbs.  S.  20°  E. ;  86  lbs.  due  E. ;  56  lbs.  N. 
33°  W. 


CONCURRENT    FORCES.  35 

2.  A  body  at  a  point  O  is  pulled  equally  by  three  men  in  direc- 
tions OA,  OB  and  0C}  such  that  AOB  =  BOC=6oa  and  AOC 
=  1200.     What  single  force  would  produce  the  same  effect  ? 

Ans.  If  P  =  pull  exerted  by  one  man,  the  resultant  is  a  force  2P 
in  the  direction  OB. 

3.  A  body  of  145  lbs.  mass  is  acted  upon  by  a  horizontal  force  of 
63  lbs.  What  single  force  would  be  equivalent  to  the  horizontal 
force  and  the  weight  of  the  body  ? 

§  2.  Moments  of  Concurrent  Forces, 

70.  Moment  of  a  Force. —  The  moment  of  a  force  with  respect  to 
a  point  is  the  product  of  the  magnitude  of  the  force  into  the  perpen- 
dicular distance  of  its  line  of  action  from  the  given  point. 

This  is  a  particular  case  of  the  definition  of  the  moment  of  a 
localized  vector,  already  given  (Art.  27).  The  sign  of  the  moment 
of  a  force  will  usually  be  assumed  to  follow  the  convention  there 
adopted. 

Moment  about  an  axis. —  A  line  through  the  origin  of  moments, 
perpendicular  to  the  plane  containing  the  origin  and  the  line  of 
action  of  the  force,  may  be  called  an  axis  of  moments.  The  moment 
of  the  force  with  respect  to  the  origin  may  also  be  regarded  as  its 
moment  with  respect  to  this  axis. 

71.  Moment  of  Resultant  of  Two  Concurrent  Forces.  —  The 
moment  of  the  resultant  of  two  concurrent  forces  with  reference  to  a 
point  in  their  plane  is  equal  to  the  algebraic  sum  of  their  separate 
moments  with  reference  to  that  point. 

Since  the  resultant  of  two  concurrent  forces  is  a  force  equal  to 
their  vector  sum  applied  at  their  common  point  of  application, 
this  proposition  is  a  particular  case  of  that  proved  in  Art.  29  for 
localized  vectors.  It  was  there  assumed  that  the  two  vectors  were 
not  parallel.  If,  however,  they  are  parallel  and  applied  at  the  same 
point,  the  truth  of  the  proposition  is  evident. 

72.  Moment  of  Resultant  of  Any  Number  of  Concurrent 
Forces. —  The  moment  of  the  resultant  of  any  number  of  concurrent 
forces  with  reference  to  any  point  in  their  plane  is  equal  to  the 
algebraic  sum  of  their  separate  moments  with  reference  to  that  point. 

The  proposition  having  been  proved  for  two  forces,  let  the  result- 
ant of  any  two  of  the  given  forces  be  combined  with  a  third ;  since  the 
proposition  is  true  for  these  two  forces  and  their  resultant,  it  is  true 


36  THEORETICAL    MECHANICS. 

for  the  three  forces  and  their  resultant.  Similarly,  it  is  true  for  this 
resultant  and  a  fourth  force,  and  is  therefore  true  for  the  four  forces 
and  their  resultant.  The  same  line  of  reasoning  may  be  extended  to 
include  any  number  of  forces. 

Examples. 

i.  Compute  the  moments  of  the  four  forces  described  in  Ex.  i, 
Art.  69,  the  origin  of  moments  being  a  point  8  ft.  directly  north  from 
the  point  of  application  of  the  forces. 

2.  Determine  the  line  of  action  of  the  resultant  of  two  forces  of 
18  lbs.  and  12  lbs.  whose  lines  of  action  include  an  angle  of  6o°. 
Compute  the  moments  of  the  two  forces  with  respect  to  an  origin  on 
the  action-line  of  the  resultant,  4  ft.  from  the  point  of  application  of 
the  forces.     Test  the  truth  of  the  above  proposition. 

3.  In  Ex.  2,  compute  the  moments  of  the  two  forces  and  their 
resultant  with  respect  to  an  origin  on  the  line  of  action  of  the  force 
of  1 2  lbs. ,  5  ft.  from  the  point  of  application. 

Ans.  77.94  ft. -lbs;  o;  77.94  ft. -lbs. 

4.  Prove  that  the  moments  of  two  concurrent  forces  are  equal 
in  magnitude  for  any  origin  on  the  line  of  action  of  the  resultant. 


§  3.  Equilibrium  of  Concurrent  Forces. 

73.  General  Condition  of  Equilibrium. —  From  the  definitions  of 
equilibrium  (Art.  57)  and  of  resultant  (Art.  55)  it  follows  imme- 
diately that  if  a  system  of  forces  is  in  equilibrium  the  resultant  is 
zero ;  and  conversely,  if  the  resultant  is  zero  the  system  is  in 
equilibrium. 

This  is  the  general  condition  of  equilibrium.      From  it  may  be 

derived     various     special     conditions, 
B  adapted  to  the  discussion  of  different 

classes  of  problems. 

74.  Geometrical  Condition  of  Equi- 
librium.—  Force  polygon.  —  The  figure 
formed  by  drawing  in  succession  vec- 
tors representing  any  number  of  forces 
in  magnitude  and  direction  is  called  a 
force  polygon  for  those  forces.  Thus, 
Fig.  25  shows  a  force  polygon  for  five 
forces  represented  by  the  vectors  AB,  BC,  CD,  DE,  EF, 


CONCURRENT    FORCES.  37 

In  general  the  initial  and  final  points  (as  A  and  F,  Fig.  25)  do 
not  coincide.  In  case  they  do  coincide  the  polygon  is  said  to  close. 
The  order  in  which  the  vectors  representing  the  forces  are  drawn 
does  not  affect  the  relative  positions  of  the  initial  and  final  points. 

Condition  of  equilibrium. —  If  A  and  F  are  the  initial  and  final 
points  of  a  force  polygon  drawn  for  any  given  forces,  AF  represents 
their  resultant  in  magnitude  and  direction.  Hence  the  resultant  will 
be  zero  if  F  coincides  with  A>  but  not  otherwise.     Therefore, 

If  any  number  of  concurrent  forces  form  a  system  in  equi- 
libriumy  their  force  polygon  is  closed.  And  conversely,  if  the  force 
polygon  closes,  the  system  is  in  equilibrium. 

75.  Moment-Condition  of  Equilibrium. —  Proposition. —  If  any 
number  of  concurrent  forces  are  in  equilibrium,  the  algebraic  sum  of 
their  moments  with  respect  to  any  point  in  their  plane  is  zero. 

It  has  been  shown  that  the  moment  of  the  resultant,  with  respect 
to  any  origin,  is  equal  to  the  algebraic  sum  of  the  moments  of  the 
components  with  respect  to  that  origin.  If  the  system  is  in  equi- 
librium the  resultant  is  zero,  therefore  its  moment  is  zero  whatever 
the  origin;  which  proves  the  proposition. 

76.  Equations  of  Equilibrium. —  From  the  principles  stated  in 
the  last  two  Articles,  any  number  of  equations  may  be  written  which 
must  be  satisfied  by  a  system  of  concurrent  forces  in  equilibrium. 
These  equations  are  of  two  kinds: 

(a)  The  sum  of  the  resolved  parts  of  the  given  forces  in  any 
direction  must  be  zero. 

For,  since  the  force  polygon  must  close  for  equilibrium,  the  alge- 
braic sum  of  the  projections  of  its  sides  upon  any  line  must  equal  zero. 

(b)  The  sum  of  the  moments  must  be  zero  for  any  origin. 
Although  each  of  these  two  general  conditions  leads  to  an  infinite 

number  of  equations,  only  two  of  these  can  be  independent.     This 
will  be  seen  by  considering  what  any  one  equation  implies. 

(1)  If  the  sum  of  the  resolved  parts  in  any  direction  is  zero,  the 
resultant  force,  if  one  exists,  must  be  perpendicular  to  that  direction. 

(2)  Jf  the  sum  of  the  moments  is  zero  with  respect  to  any  point, 
the  resultant  force,  if  one  exists,  must  act  in  a  line  passing  through 
that  point. 

It  is  now  evident  that  there  will  be  equilibrium  if  either  of  the 
following  conditions  is  satisfied: 


38  THEORETICAL    MECHANICS. 

I.  The  sum  of  the  resolved  parts  of  the  forces  is  zero  in  each  of 
two  directions. 

II.  The  sum  of  the  moments  is  zero  with  respect  to  each  of  two 
points  not  collinear  with  the  point  of  application  of  the  forces. 

III.  The  sum  of  the  moments  is  zero  for  one  origin  (not  coincid- 
ing with  the  point  of  application  of  the  forces),  and  the  sum  of  the 
resolved  parts  is  zero  for  any  one  direction  not  perpendicular  to  the 
line  joining  the  origin  of  moments  with  the  point  of  application. 

77.   Algebraic    Deduction   of    Conditions    of     Equilibrium. — 

With  the  notation  of  Art.  69,  we  have 

X  =  Px  cos  ax  -\-  P2  cos  a  ,  -f     .      .      .      , 
K==Plcos/8I  +  />cos0,  +     .     .     .     , 


R  =  XX1  +  Y\ 

The  general  condition  of  equilibrium,  R  ±=  o,  requires  that 

X  =  o   and  Y  ==  o. 

For  unless  X 2  and  Y1  are  separately  equal  to  zero,  their  sum  cannot 
equal  zero  unless  one  of  them  is  negative;  but  this  would  make  X 
or  Y  imaginary.  It  follows  that  the  condition  ft  ==  o  is  equivalent 
to  the  two  conditions 

Px  cos  ax  -j-  P2  cos  a%  +     .     .     .     =  o ;         .       ( 1 ) 

Px  cos  j3x  +  Pt  cos  /3,  +     ...     =0.  .       (2) 

Since  the  axes  along  which  the  forces  are  resolved  may  be  any 
pair  of  rectangular  axes  in  the  plane  of  the  forces,  it  follows  that  for 
equilibrium  the  sum  of  the  resolved  parts  of  the  forces  in  any  direc- 
tion must  be  zero.     This  agrees  with  Art.  76. 

It  might  be  shown  algebraically  that  if  two  equations  such  as  (1) 
and  (2)  are  satisfied,  every  equation  obtained  by  resolving  the  forces 
in  any  direction  or  by  taking  their  moments  about  any  origin  must 
also  be  satisfied. 

78.  Solution  of  Problems  in  Equilibrium. —  In  solving  prob- 
lems in  equilibrium  algebraically,  it  is  necessary  to  write  as  many 
independent  equations  as  there  are  unknown  quantities  to  be  deter- 
mined. From  the  above  principles  of  equilibrium  for  concurrent 
forces,  two  independent  equations  can  always  be  written.      If  the 


CONCURRENT    FORCES. 


39 


number  of  unknown  quantities  is  greater  than  two,  the  principles  of 
equilibrium  alone  are  not  sufficient  for  their  determination.  In  such 
a  case  the  problem,  if  determinate,  involves  something  more  than  the 
principles  of  equilibrium. 

From  the  preceding  discussion  it  is  evident  that  the  two  inde- 
pendent statical  equations  can  always  be  written  with  certainty.  If 
the  solution  of  the  problem  presents  difficulty  it  lies  usually  in  the 
algebraic  solution  of  the  equations. 

In  writing  the  two  independent  equations  of  equilibrium  in  either 
of  the  three  allowable  ways  (Art.  76),  simplification  is  usually  pos- 
sible by  taking  advantage  of  the  following  obvious  principles : 

(a)  If  the  resolution  is  made  in  a  direction  perpendicular  to  a 
force,  that  force  does  not  enter  the  resulting  equation. 

(b)  If  moments  are  taken  about  an  origin  lying  on  the  line  of 
action  of  a  force,  that  force  does  not  enter  the  resulting  equation. 

In  many  cases  the  ''geometrical  condition  of  equilibrium,"  that 
the  force  polygon  must  close,  leads  to  a  short  solution  without  the 
use  of  the  equations  of  equilibrium. 

It  is  thus  seen  that  any  problem  may  be  analyzed  in  two  ways, — 
geometrically  and  algebraically.  In  solving  the  examples  that  fol- 
low, the  student  is  advised  to  use  both  methods.  The  geometrical 
method  frequently  has  the  advantage  of  giving  a  rapid  solution, 
while  it  also  gives  a  clear  view  of  the  relations  of  the  forces.  On 
the  other  hand,  the  algebraic  method  has  the  advantage  of  affording 
an  exhaustive  treatment  applicable  to  any  determinate  problem. 

79.  Applications. —  The  methods  of  applying  the  foregoing  prin- 
ciples will  be  illustrated  by  the  solution  of  the  following  problems. 

I.  A  body  of  40  lbs.  mass  is  suspended  by  a  cord  from  a  ring 
which  is  supported  by  two  cords  making  angles  of  30°  and  70°  with 
the  vertical.     Determine  the  tensions  in  the  cords.* 

Two  systems  of  forces  in  equilibrium  are  presented  here, — (a) 
two  forces  acting  upon  the  suspended  body  (its  weight  of  40  lbs. 
directed  downward  and  an  upward  force  exerted  by  the  string),  and 
(b)  three  forces  acting  upon  the  ring  (exerted  by  the  three  cords). 
The  conditions  of  equilibrium  are  to  be  applied  to  each  system  sep- 
arately. 

*  In  all  cases  in  which  cords  are  introduced  they  are  understood  to  be 
perfectly  flexible.     (Art.  43.) 


4° 


THEORETICAL    MECHANICS. 


(a)  The  upward  force  exerted  by  the  cord  CD  upon  the  body  D 
(Fig.  26)  must  equal  the  downward  force  of  40  lbs.  due  to  gravity. 

(b)  The  cord  CD  must  pull  down- 
ward upon  the  ring  C  with  a  force  equal 
to  its  upward  pull  upon  D.  Hence,  of 
the  three  forces  acting  upon  the  ring, 
one  is  known  completely  and  the  direc- 
tions of  the  other  two  are  known. 

Geometrical  solution. —  The  vectors 

representing  the  three  forces  applied  to 

the  ring  must  form  a  closed  triangle. 

Draw  LM(Fig.  26)  vertically  downward 

to  represent  the  tension  in    CD,  MN 

parallel    to    CB,   LN  parallel   to    AC', 

then  MN  and  NL  must  represent  the 

forces  exerted  upon  the  ring  by  the  cords  BC  and  AC  respectively. 

From  the  given  data,  LM=  40  lbs.,  angle  LMN=  30°,  angle 

MLN '-—  70°,  angle  LNM  —  8o°.      Hence,  by  trigonometry, 


Fig.  26. 


LN 

sin  3oc 


or        LN  —  LM 


sin  30 


MN 
sin  7oc 

a  5 


sin  8o°        0.9848 


LM 

sin  8oc 


X  40  lbs.  =  20. 3  lbs. ; 


MN  =  LM 


sin  70         Q-9397 
sin  8o°  ~~  0.9848 


X  40  lbs.  =  38. 2  lbs. 


Algebraic  solution. —  Let  P  and  Q  denote  the  tensions  in  AC 
and  CB,  and  let  the  independent  equations  of  equilibrium  be  formed 
by  resolving  forces  horizontally  and  vertically.  Then  (taking  up- 
ward and  toward  the  right  as  positive  directions) 

—  P  cos  20°  -f-  Q  cos  6o°  =  o  ; 

P  cos  700  -j-  Q  cos  30°  —  40  lbs.  ==  o. 

Or  —0.9397  P-f  0.5000  2  =  o; 

o.  3420  P  -f-  o.  8660  Q  =  40  lbs. 
Solving,  P=  20.3  lbs.,  Q=  38.2  lbs. 


CONCURRENT    FORCES.  41 

II.  A  body  of  60  lbs.  mass  rests  against  a  smooth  plane  surface 
inclined  25°  to  the  horizontal.  It  is  supported  partly  by  the  plane 
and  partly  by  a  cord  inclined  20°  to  the  plane.  What  are  the  magni- 
tudes of  the  supporting  forces  ?     (  Fig. 

The  forces  acting  upon  the  body  are                            ^^/^^^ 
three,  all  known  in  direction  (the  weight                      ^-^^^ 
acting  vertically  downward,  the  pull  of         ^^^^^5° 
the  cord  along  its  length,  and  the  pres-      ^T_ i 

sure  of  the  smooth  plane  normally  to  its    .  Fig.  27. 

surface),  and  one  known  in  magnitude. 

Geometrical  solution. —  The  triangle  of  forces  can  be  drawn  and 
the  unknown  forces  determined  as  in  the  preceding  problem. 

Algebraic  solution. —  In  writing  the  equations  of  equilibrium,  it 
will  be  advantageous  to  resolve  forces  parallel  and  perpendicular  to 
the  plane.     This  gives  an  equation  not  involving  the  normal  pressure. 

Let  P  =  tension  in  cord,  and  Q  =  pressure  exerted  by  plane. 

Resolving  parallel  to  the  plane, 

P  cos  200  —  60  cos  6 50  =  0. 
Resolving  perpendicularly  to  the  plane, 

P  cos  700  -{-  Q  —  60  cos  250  =  o. 
Solving,  P  =  27.  o  lbs. ,  Q  =  45. 1  lbs. 

Examples. 

1.  A  body  of  20  lbs.  mass  resting  upon  a  horizontal  surface  is 
acted  upon  by  gravity  and  by  a  horizontal  force  of  5  lbs.  The  sup- 
porting body  exerts  upon  it  such  a  force  as  to  hold  it  at  rest. 
Determine  the  magnitude  and  direction  of  this  force. 

Ans.  20.6  lbs.,  inclined  14°  2  to  the  normal. 

2.  A  body  of  120  lbs.  mass,  suspended  by  a  flexible  string,  is 
pulled  horizontally  with  a  force  of  45  lbs.     Determine  the  direction 

of  the  suspending  cord  and  the  tension  sus- 
tained by  it.     (Fig.  28.) 

3.  A  body  of  70  lbs.  mass  is  suspended  by 
a  flexible  cord  from  a  ring  which  is  supported 
by  two  cords  making  angles  of  25°  and  6o° 
respectively  with  the  vertical.     Determine  the 

>■     tensions  in  the  strings. 

Fig.  28.  4.   A  body  of  P  lbs.  mass  is  suspended  by 


42 


THEORETICAL    MECHANICS. 


a  cord  from  a  ring  which  is  supported  by  two  cords  making  angles 
a  and  fi  with  the  vertical.     Determine  the  tensions  in  the  cords. 
Ans.  Psm  a  /sin  (a  -\-  P)  and  P  sin  /3/sin  (a  -f  /3). 

5.  A  body  of  70  lbs.  mass  is  suspended  as  in  Ex.  3,  and  is  also 
pulled  horizontally  by  means  of  an  at- 
tached cord  with  a  force  of  10  lbs.     De- 
termine the  tension  in  each  cord.     (Fig. 
29.) 

[This  presents  two  sets  of  concurrent 
forces,  one  set  acting  on  the  suspended 
body,  the  other  acting  on  the  ring.] 

6.  If  the  horizontal  force  in  Ex.  5  be 
gradually  increased,  one  of  the  cords 
will  finally  cease  to  act.  For  what  value 
of  the  horizontal  force  will  this  occur  ? 

In  this  limiting  case,  what  tensions  are 
sustained  by  the  cords  ? 

Ans.  Horizontal  force  =  121.24 
lbs.     Tensions  =140  lbs.  and  o. 

7.  In  Fig.  30  the  cords  a,  b,  c  and 
d  are  inclined  to  the  vertical  at  angles 
of  70°,  30°,  20°  and  o°  respectively, 
while  fy  g  and  h  are  horizontal.  The 
mass  of  the  body  P  is  60  lbs. ,  and  the 
horizontal  force  exerted  by  the  cord  h 
is  20  lbs.  Determine  the  tension  in 
each  of  the  cords. 

[Apply  conditions  of  equilibrium 
separately  to  the  forces  concurrent  at 
Z,  at  M,  at  N  and  at  P,  beginning 
with  P.] 

Ans.  Tension  in  a  =  49.7  lbs. ;  in 
c,  63.8  lbs.;  in  g,  20  lbs. 

80.  Equilibrium  of  a  Particle  on  a  Smooth  Surface.  —  If  a 
particle  rests  against  a  smooth  surface  of  any  form,  the  pressure  ex- 
erted upon  the  particle  by  the  surface  is  a  "passive  resistance" 
(Art.  41)  whose  direction  is  known,  being  that  of  the  normal  to  the 
surface.  If  the  surface  is  a  plane,  the  direction  of  the  pressure  is  the 
same  whatever  the  position  of  the  particle  ;  but  for  a  curved  surface 
the  direction  of  the  pressure  exerted  by  the  surface  depends  upon 
the  position. 

To  solve  a  problem  relating  to  the  equilibrium  of  a  particle  on  a 
smooth  surface,  the  general  equations  of  equilibrium  may  be  written 
in  the  usual  manner  ;  but  among  the  forces  to  be  included  in  the 


CONCURRENT    FORCES. 


43 


o 


system  is  the  unknown  pressure  due  to  the  surface.     The  direction 
of  this  reaction  is  known  if  the  position  of  the  particle  is  known  ; 
otherwise  it  must  be  expressed  in  terms  of  the  variable  coordinates 
of  the  surface.     The  consideration  of 
this   problem  will  be  confined  to   the 
case  in  which  the  particle  and  the  ap- 
plied forces  are  restricted  to  a  plane 
curve  lying  in  the  given  surface. 

Let  the  particle  be  restricted  to 
the  plane  curve  shown  in  Fig.  31, 
and  let  the  equation  of  this  curve  re- 
ferred to  rectangular  axes  OX,  OY 
be  known.  If  N  denotes  the  mag- 
nitude of  the  unknown  normal  pressure  acting  on  the  particle,  and  cf> 
the  angle  between  the  ^r-axis  and  the  direction  of  N>  we  have 

^•-component  of  JV—  iVcos  </>  ; 

j-component  of  N  =  —  N  sin  (/>. 

But  since  N  has  the  direction  of  the  normal  to  the  curve, 

dy_  dyjdx 


Fig.  31. 


COS  (j) 


sin  cj> 


ds       Vi  4.  (dy/dxf 
dx  1 


ds      1 7 1  -f-  (dy/dx)1 


The  value  of  dyjdx  in  terms  of  x  and  y  can  be  found  from  the  equa- 
tion of  the  curve. 

In  many  cases  the  direction-angles  of  N  can  be  expressed  more 
conveniently  in  some  other  manner. 

The  method  of  solving  problems  of  this  kind  will  now  be  illus- 
trated. 

I.  A  smooth  wire  bent  into  the  form  of  a  circle  and  placed  with 
a  diameter  vertical  carries  a  bead  of  W  lbs.  mass,  to  which  is  applied 
a  horizontal  force  of  P  lbs.  In  what  position  can  the  bead  be  in 
equilibrium  ? 

Algebraic  solution. —  Let  the  radius  drawn  to  the  bead  make  with 
the  vertical  an  angle  0,  and  let  N  denote  the  pressure  of  the  wire 
upon  the  bead,  its  direction  being  radial.     (Fig.  32.) 


44 


THEORETICAL    MECHANICS. 


Resolving  forces  horizontally, 

P—  N  sin  0  =  o. 


Resolving  vertically, 


W—Ncos0  =  o. 


Solving    these   equations   for    the    unknown    quantities   N  and    0, 

we  have 

tan0  =P/W; 


Fig.  32. 


N  =  P/sin  6  =  V  P*  +  W\ 

Geometrical  solution. — The  par- 
ticle is  in  equilibrium  under  the  action 
of  three  forces,  P,  W  and  N.  Two 
of  these  (P  and  W")  are  known  ; 
hence  the  force  triangle  can  be  drawn 
completely,  thus  determining  the  mag- 
nitude and  direction  of  N.  From  the 
triangle  it  is  evident  that  N  acts 
toward  the  center  ;  that  its  magni- 
tude is  VP2  +   W2 ;  and  that  tan  0  =  P/W. 

II.  A  particle  whose  mass  is  Wlbs.  rests  upon  a  smooth  hori- 
zontal plane  ;  to  it  are  attached  a  string 
AB  passing  over  a  smooth  peg  at  B  and 
sustaining  a  body  of  mass  P  lbs. ,  and  a 
string  A  C  passing  over  a  smooth  peg  at 
C  and  sustaining  a  body  of  mass  Q  lbs. 
Required  the  direction  of  AC  for  equi- 
librium. 

Algebraic  solution.  —  The  particle  is 
acted  upon  by  four  forces  :  its  weight, 
W  lbs. ;   a  force  of  P  lbs.  due  to  the 

tension  in  the  string  AB  ;    a  force  of  Q    lbs.  due  to  the  tension 
in  the  string  A  C ;  and  the  normal  pressure  N  due  to  the  plane. 

Let  0  denote  the  angle  between  A  C  and  the  horizontal.     Re- 
solving forces  horizontally  and  vertically, 

—  P  +  Q  cos  0  =  o  ; 


N  —  W  +  Q  sin  0  =  o. 


CONCURRENT    FORCES. 


45 


Solving  for  N  and  0, 


cos  6  =  P/Q ; 


N=  W—VQ*  —  P\ 


The  geometrical  solution  by  the  force  triangle  is  obvious. 

III.  To  determine  the  position  of  equilibrium  of  a  body  resting 
on  the  surface  of  a  smooth  elliptic  cylinder  and  acted  upon  by  a  force 
of  known  magnitude  directed  along  the  tangent  to  the  ellipse. 

Let  the  mass  of  the  body  be  W  lbs. ,  and  let  the  tangential  force 
be  due  to  the  tension  in  a  string  which  passes  over  a  smooth  peg  and 
sustains  a  body  of  known  mass  P 
lbs.,  as  in  Fig.  34. 

The  body  is  acted  upon  by  three 
forces  :  its  weight  W  lbs.  acting 
vertically  downward  ;  the  pull  of  the 
cord,  its  magnitude  being  P  lbs.  and 
its  direction  that  of  the  tangent  to 
the  ellipse  ;  and  the  normal  pres- 
sure exerted  by  the  smooth  surface, 
its  magnitude  N  being  unknown. 

If  (f>  denotes  the  angle  between  the  normal  and  the  axis  of  x, 
measured  as  shown  in  the  figure,  the  equations  obtained  by  resolving 
along  the  tangent  and  normal  are 


Fig.  34. 


From  (1), 
From  (2), 


W cos  </>  —  P=o; 
N  —  W  sin  (j>  =  o. 
cos  <f>  =  PI  W. 


N=  Wmt=y  W*  —  Px 


(0 
(2) 

(3) 


The  coordinates  of  the  point  for  which  cos  <b  —  Pj  W  may  be 
found  as  follows  : 

dyjdx 


cos  <f>  = —  - 

ds 

The  equation  of  the  ellipse  is 


V  1  -f  {dyfdxf 


y 


=  I. 


(4) 


46 

THEORETICAL 

MECHANICS. 

Differentiating, 

dy 
dx 

2         ' 

ay 

which  substituted 

in  the  value  of  cos  <f>  gives 

&2x 

P 

cos  <f>  =  —  —  -  =  — .  .         •     (5) 

Va'y'1  +  b\xl        W 

The  coordinates  of  the  position  of  equilibrium  must  satisfy  equations 
(4)  and  (5).     Solving, 


?—.     / Plal . 

«       ^ P'lal  -I-  (#"  —  /")£,; 

£  =     /~     (H^2-P2)£2 

^        >/>2tf2  +  (W*  —  P*)62' 


Examples. 

1.  A  body  of  Wlbs.  mass,  resting  on  a  smooth  plane  surface 
inclined  at  angle  6  to  the  horizontal,  is  supported  by  the  pressure  of 
the  plane  and  another  force  acting  at  angle  a  with  the  plane.  De- 
termine the  supporting  forces. 

Ans.  Normal  pressure  =  W  cos  (6  -\-  a) /cos  a. 

2.  In  what  direction  must  a  force  of  P  lbs.  be  applied  to  a  body 
whose  mass  is  W  lbs.  to  hold  it  at  rest  on  a  smooth  plane  inclined 
at  angle  a  to  the  horizontal  ? 

Ans.  If  6  =  angle  between  Pand  the  plane,  cos  6  =  (  IV/P)  sin  a. 

3.  A  certain  body  may  be  supported  on  a  smooth  inclined  plane 
by  a  force  P  acting  at  an  angle  a  with  the  horizontal,  or  by  a  force 
Q  acting  at  an  angle  /3  with  the  horizontal.  Required  the  mass  W 
of  the  body  and  the  inclination  6  of  the  plane  to  the  horizontal. 

Ans.X*ne  =  Pc0%a-Qc0sli.     W=     PQ  *™  W  ~  a)    . 

Q  sin  j3  —  P  sin  a  P  cos  a  —  (2  cos  /3 

4.  A  certain  body  may  be  supported  on  a  smooth  inclined  plane 
by  a  horizontal  force  of  10  lbs.,  or  by  a  force  of  8  lbs.  acting  along 
the  plane.  Required  the  mass  of  the  body  and  the  inclination  of 
the  plane.  Ans.  13^  lbs.;  360  52'. 

5.  A  body  whose  mass  is  10  lbs.  is  supported  on  a  smooth  in- 
clined plane  by  a  force  of  2  lbs.  acting  along  the  plane  and  a  hori- 
zontal force  of  5  lbs.     Determine  the  inclination  of  the  plane. 

6.  A  smooth  wire,  bent  into  the  form  of  a  parabola  with  axis  ver- 
tical and  vertex  downward,  carries  a  bead  of  mass  W  lbs.  which  is 


CONCURRENT    FORCES. 


47 


pulled  horizontally  by  a  force  of  P  lbs.     Determine  the  position  of 
equilibrium. 

Ans.  If  the  equation  of  the  parabola  is  x2  =  qmy,  the  coordi- 
nates of  the    position   of  equilibrium  are  x  =   (2PI  W)m,  y  = 

(pyw^m. 

7.  A  heavy  bead  is  placed  upon  a  smooth  circular  wire  in  a  vertical 
plane,  and  is  pulled  by  a  string  which  passes  over  a  smooth  pulley 
at  the  highest  point  of  the  circle  and  carries  a  body  of  known  weight. 
Find  the  position  of  equilibrium. 

8.  Two  particles  whose  masses  are  P  lbs.  and  Q  lbs.  are  con- 
nected by  a  flexible  string  of  length  /  which  passes  over  a  smooth 
circular  cylinder  of  diameter  d  with  axis 

horizontal.      Required   the   position   of 
equilibrium.     (Fig.  35.) 

[Apply  the  conditions  of  equilibrium 
to  each  particle  separately,  remembering 
that  the  tension  in  the  string  is  uniform 
throughout  its  length,  so  that  it  exerts 
equal  forces  on  the  two  particles.  Both 
particles,  or  only  one  of  them,  may  be 
in  contact  with  the  cylinder  in  the  posi- 
tion of  equilibrium ;  these  two  cases 
must  be  treated  separately.] 

9.  In  the  case  described  in  Ex.  8, 
let  the  masses  of  the  particles  be  2  lbs. 

and  3  lbs. ,  the  length  of  the  string  1  ft. , 
and  the  diameter  of  the  cylinder  18  ins. 
Determine  the  position  of  equilibrium, 
the  tension  in  the  string,  and  the  reac- 
tions exerted  on  the  particles  by  the 
cylinder. 

10.  A  string  attached  to  a  fixed 
point  A  and  passing  over  a  smooth  peg 
B,  carries  at  the  free  end  a  body  of 
known  weight  P,  and  a  body  of  weight 
W  is  suspended  at  a  point  C  by  means 
of  another  string.  The  length  A  C  and 
the  positions  A  and  B  being  known,  it 
is  required  to  determine  the  position  of  equilibrium.  (Fig.  36.) 
[The  solution  involves  an  equation  of  the  third  degree.] 

1 1.  In  Ex.  10,  let  the  weight  W  be  suspended  from  a  smooth  ring 
sliding  on  the  string  A  CB.     Determine  the  position  of  equilibrium. 

12.  Given  three  concurrent  forces  of  magnitudes  Py  Q,  R,  the 
angles  between  P  and  Q,  Q  and  R,  R  and  P,  respectively,  being  rt 
py  q.     Prove  that  the  square  of  the  resultant  is  equal  to 

P'1  +  Q*+  R2  +  2PQ  cos  r  +  2QR  cos/  +  2RPcos  q. 


Fig.  35. 


WO 


Fig.  36. 


THEORETICAL    MECHANICS. 


13.  AB  and  AC  (Fig.  37)  are  smooth  rods  lying  in  a  vertical 
plane.  Heavy  rings  of  known  masses  P  lbs.  and  Q  lbs.  slide  upon 
the  rods,  being  connected  by  a  flexible  string  MN.     Required  the 

inclination  of  MN  to  the  horizontal 
in  the  position  of  equilibrium. 

14.  Two  particles  whose  masses 
are  P  lbs.  and  Q  lbs.  respectively  rest 
against  smooth  inclined  planes  whose 
inclinations  to  the  horizontal  are  a 
and  /3.  The  particles  are  connected 
by  a  flexible  cord  which  passes  over 
a  smooth  peg  placed  vertically  above 
If/  is 


Fig.  37. 


the  intersection  of  the  planes, 
the  length  of  the  cord  and  h  the  ver- 
tical distance  of  the  peg  above  the  in- 
tersection of  the  planes,  prove  that  the 
angles  6  and  <f>  made  by  the  cord  with 
the  planes  in  the  position  of  equilibrium 
are  determined  by  the  equations 

p  sin  a n  sin  /3     cos  a       cos  j3  __/ 

cos  0  cos  (j)     sin  6       sin  </>      h 

15.  Determine  the  magnitude  and 
direction  of  the  least  force  that  will 
sustain  a  particle  of  mass  W  kilograms 

on  a  smooth  plane  inclined  at  an  angle  a  with  the  horizon.     Com- 
pute results  if  W  =  45  and  a  =  360. 

16.  A  particle  on  a  smooth  plane  inclined  at  angle  a  to  the 
horizon  is  acted  on  by  a  force  directed  toward  a  fixed  point  and 
varying  inversely  as  the  square  of  the  distance  from  that  point.  De- 
termine the  position  of  equilibrium. 

17.  Two  smooth  rings  of  14  lbs.  and  18  lbs.,  connected  by  a  flex- 
ible string  2  ft.  long,  slide  on  a  smooth  vertical  circular  wire  of  radius 
5  ft.      Determine  the  position  of  equilibrium. 

Arts.  Angle  between  string  and  horizontal  =  i°  28'. 


CHAPTER    IV. 

COMPOSITION    AND    RESOLUTION    OF    NON-CONCURRENT    FORCES    IN 
THE   SAME   PLANE. 

§  i.  Two  Non-Concurrent  Forces. 

81.  Rigid  Body. —  In  the  discussions  which  follow,  the  systems 
of  forces  are  in  most  cases  regarded  as  applied  to  a  rigid  body. 

A  rigid  body  may  be  defined  as  one  whose  particles  do  not  change 
their  positions  relative  to  one  another  under  any  applied  forces.  No 
known  body  satisfies  this  condition  strictly,  even  for  forces  of  small 
magnitude,  but  in  the  solution  of  problems  of  practical  importance 
most  solid  bodies  may,  with  slight  error,  be  regarded  as  rigid.  After 
a  body  has  assumed  a  form  of  equilibrium  under  applied  forces  it 
may,  in  applying  the  principles  of  Statics,  be  treated  as  a  rigid  body 
without  error. 

82.  Change  of  Point  of  Application. —  The  effect  of  a  force  upon 
the  motion  of  a  rigid  body  will  be  the  same,  at  whatever  point  in  its 
line  of  action  it  is  applied,  if  the  particle  upon  which  it  acts  is  rigidly 
connected  with  the  body. 

This  proposition,  which  is  amply  justified  by  experience,  is  funda- 
mental to  the  development  of  the  principles  of  Statics.  In  applying 
the  principle,  we  are  at  liberty  to  assume  a  point  of  application  outside 
the  actual  body,  the  latter  being  ideally  extended  to  any  desired 
limits. 

83.  Collinear  Forces. —  Two  forces  having  the  same  line  of 
action,  applied  to  the  same  rigid  body  at  any  points  of  that  line,  may 
be  combined  as  if  they  were  concurrent ;  since  by  Art.  82  both  may 
be  treated  as  if  applied  at  any  one  point  in  their  common  line  of 
action. 

As  a  particular  case,  two  forces  applied  to  a  rigid  body  balance 
each  other  if  they  are  equal  in  magnitude,  opposite  in  direction,  and 
collinear.  Hence  'such  a  pair  of  forces  may  be  introduced  into  a 
system  without  changing  its  effect. 

84.  Resultant  of  Two  Non-Parallel  Forces. —  If  two  coplanar 
forces  are  not  parallel,  their  lines  of  action  must  intersect,  and  the 

4 


50  THEORETICAL    MECHANICS. 

point  of  intersection  may  be  treated  as  their  common  point  of  appli- 
cation. They  may  therefore  be  treated  as  concurrent  forces,  and 
their  resultant  may  be  determined  as  in  Art.  59.  Hence  the  following 
proposition  may  be  stated: 

The  resultant  of  two  non-parallel  forces  acting  in  the  same  plane 
on  a  rigid  body  is  a  force  equal  to  their  vector  sum,  and  its  line  of 
action  passes  through  the  point  of  intersection  of  the  lines  of  action 
of  the  given  forces.  Its  point  of  application  may  be  any  point  of 
this  line. 

Thus,  if  two  forces  are  applied  to  a  body  at  points  M  and  N 
(Fig.  39),  their  lines  of  action  being  MS  and  NT,  each  may  be 

assumed  to  act  at  R,  the  point  of  inter- 
section of  these  lines.  The  resultant 
must  therefore  be  a  force  which  may 
be  taken  as  acting  at  R  ;  that  is,  its 
line  of  action  must  pass  through  R. 
When  the  magnitude  and  direction  of 
the  resultant  have  been  found  by  con- 
structing the  vector  triangle,  the  line  of 
action  becomes  known,  and  any  point  of  this  line  may  be  regarded 
as  the  point  of  application. 

By  an  extension  of  the  above  reasoning,  any  number  of  forces 
whose  lines  of  action  meet  in  a  point 

may  be  treated  as  if  that  point  were  7> 

their  common  point  of  application;  in  y'l  \ 

other  words,  as  if  they  were  concur-  /    , 

rent.     This  is  true   even  though  the  /        j 

point   of  intersection    of  the   lines   of        'aJl. / \C 

action   falls  outside  the  limits   of  the  A ^5 V 

body.     Thus,  if  the  forces  Plt  Pti  P3     /p  1  p\ 

are  applied  to   the  bar  AC  (Fig.  40)  *  O 

at   points  A,    B   and    C,    their  direc-  Fig.  40. 

tions   being   such    that    their   lines   of 

action  intersect  at  a  point  D,  their  effect  is  the  same  as  if  all  were 

applied  at  D}  the  latter  point  being  regarded  as  rigidly  connected 

with  the  bar. 

85.  Resolution  of  a  Force  Into  Two  Non-Parallel  Compo- 
nents.—  Let  M  (Fig.  41)  be  the  point  of  application  of  a  force,  and 
MS  its   line   of  action.     Any   point  in   this   line,  as   R,  may  be 


COMPOSITION    OF    NON-CONCURRENT    FORCES. 


51 


regarded  as  the  point  of  application,  and  the  force  may  be  replaced  by- 
two  components  acting  at  R.     The  magnitudes  and  directions  of  the 
two  components  must  be  such  that  their 
vector  sum  is  equal  to  the  given  force. 

86.  Resultant  of  Two  Parallel 
Forces. —  In  determining  the  resultant 
of  two  parallel  forces,  two  cases  must  be 
considered,  according  as  the  forces  act 
in  the  same  direction  or  in  opposite 
directions. 

(1)  Forces  having  the  same  direc- 
tion. —  Let  P  and  Q  denote  the  magnitudes  of  two  forces  having 
the  same  direction,  A  and  B  (Fig.  42)  being  any  points  in   their 
lines  of  action.     Let  two  forces,  each  of  magnitude  F,  be  applied 


Fig.  41. 


tRm 


Fig.  42. 


to  the  body,  acting  in  opposite  directions  along  the  line  AB; 
call  these  forces  F'  and  F" .  The  system  of  four  forces  (P,  Q, 
F\  F")  is  equivalent  to  the  given  system  of  two  forces  (P,  Q)t 
since  F'  and  F"  balance  each  other.  Let  the  force  F'  be  com- 
bined with  P  (giving  a  resultant  R)  and  the  force  F"  with  Q 
(giving  a  resultant  R").  The  line  of  action  of  R'  passes  through 
A,  and  its  direction  is  that  of  the  diagonal  of  a  parallelogram  with 
sides  drawn  from  A,  proportional  and  parallel  to  P  and  F.  The 
line  of  action  of  R"  passes  through  B,  and  is  found  by  construct- 
ing a  parallelogram  with  sides  drawn  from  B,  proportional  and 
parallel  to  Q  and  F.  The  forces  R  and  R'  may  be  regarded  as 
applied  at  D,  the  intersection  of  their  lines  of  action ;  R  is  therefore 


52  THEORETICAL   MECHANICS. 

equivalent  to  two  forces  equal  and  parallel  to  P  and  F'  applied  at 
D,  and  R"  is  equivalent  to  two  forces  equal  and  parallel  to  Q  and 
F"  applied  at  D;  these  four  forces  applied  at  D  being  thus  equiv- 
alent to  the  given  forces  P  and  Q.  But  since  the  forces  F'  and  F'' 
balance  each  other,  the  system  is  equivalent  to  two  collinear  forces  of 
magnitudes  P  and  Q,  or  to  a  single  force  P  -f  Q,  acting  at  D  parallel 
to  the  given  forces.  This  single  force  is  therefore  the  resultant  of 
the  two  given  forces. 

It  remains  to  determine  the  position  of  the  line  of  action  of  this 
resultant  relative  to  the  lines  of  action  of  the  given  forces.  Let  C 
be  the  point  in  which  this  line  intersects  AB;  then  by  similar  tri- 
angles, 

F/P  =  AC/ CD,  and  F/Q  =  CB/CD. 

Combining  these  two  equations, 

AC/CB  -=  Q/P; 

that  is,  the  line  of  action  of  the  resultant  divides  AB  into  segments 
inversely  proportional  to  Pa.nd  Q. 

(2)  Forces  having  opposite  directions. —  The  case  in  which  P  and 
Q  have  opposite  directions  is  represented  in  Fig.  43  ;  the  reasoning 

r;      p 


/b 


D 
R 


Fig.  43. 


employed  in  the  preceding  case  applies  equally  to  this,  the  two  fig- 
ures being  lettered  in  a  corresponding  manner.  The  magnitude  of 
the  resultant  is  found  to  be  the  difference  (or  algebraic  sum)  of  P 
and  Q.  Also,  the  point  D  and  therefore  the  line  CD  will  fall  outside 
the  space  included  between  the  lines  of  action  of  P  and  Q  and  on  the 
side  of  the  greater  of  these  forces.  In  this  case  the  point  C  divides 
AB  externally  into  segments  inversely  proportional  to  P  and  Q. 


COMPOSITION    OF   NON-CONCURRENT    FORCES.  53 

The  examples  which  follow  are  designed  to  illustrate  the  method 
of  finding  the  resultant  of  two  parallel  forces,  or  of  finding  two  par- 
allel forces  that  are  equivalent  to  a  single  force.  The  reader  who  is 
already  acquainted  with  the  principles  of  equilibrium  (to  be  developed 
later)  will  notice  that  several  of  the  examples  can  be  solved  by  means 
of  these  principles.  It  is,  however,  desirable  that  they  should  be 
analyzed  by  the  principles  of  composition  and  resolution  of  forces 
already  developed;  the  resultant  of  two  forces  being  regarded  as  a 
single  force  which  may  replace  them  without  affecting  the  state  of 
the  body  as  regards  rest  or  motion. 

Thus,  in  example  3  of  the  following  list,  the  bar  AB  may  be  acted 
upon  by  any  number  of  forces  not  specified;  these,  together  with  the 
two  upward  forces  exerted  by  the  strings,  maintain  the  bar  in  its 
condition  of  rest.  Now  suppose  the  action  of  these  two  forces  ceases; 
what  single  force  may  be  applied  with  the  same  result  ? 


Examples. 

1.  Parallel  forces  of  40  lbs.  and  65  lbs.  acting  in  the  same  direc- 
tion are  applied  at  the  ends  of  a  bar  1 2  ft.  long.  Find  the  magnitude, 
direction  and  line  of  action  of  a  single  force  which  would  produce  the 
same  effect. 

2.  Assuming  the  two  forces  in  the  preceding  example  to  act  in 
opposite  directions,  determine  their  resultant  completely. 

3.  A  heavy  bar  AB  (Fig.  44),  8  ft.  long,  is  supported  by  vertical 
cords  attached  at  A  and  B,  which 

pass  over  smooth  pegs  at  C  and  D       Q  C  D 

and  sustain  bodies  P  and  Q  of  masses 

100  lbs.  and  60  lbs.  respectively.     If 

the  two  supporting  cords  are  replaced 

by  a  single  cord,  at  what  point  must 

it  be  attached,  and  what  force  must  it 

exert?  D?  QO 

4.  A  heavy  bar  AB,  12  ft.  long,  Fig.  44. 
is  found  to  balance  on  a  smooth  sup- 
port at  C,  7.5  ft.  from  one  end,  the  upward  pressure  exerted  by  the 
support  being  48  lbs.     If  the  same  bar  rests  on  two  smooth  supports 
at  the  ends,  what  will  be  the  supporting  forces  ? 

5.  A  heavy  bar  18  ft.  long  is  supported  at  the  ends  by  equal 
vertical  forces  of  40  lbs.  If  the  supports  are  moved  to  points  4  ft. 
from  one  end  and  6  ft.  from  the  other  respectively,  what  are  the  sup- 
porting forces  ? 


54  THEORETICAL    MECHANICS. 

6.  The  bar  AB  (Fig.  45)  carries  a  load  P  of  20  lbs.  hanging 
by  a  string  from  the  end  A,  and  is  supported  by  a  string  attached 

at  C,  4  ft.  from  A,  the  tension  of  the 
supporting  string  being  found  by  a 
spring  balance  to  be  40  lbs.  If  the 
load  P  is  removed,  at  what  point  must 
the  string  be  attached  in  order  to 
support  the  bar  ?     What  tension  will 

B      it  sustain  ? 

I  Ans.   8  ft.  from  A;  20  lbs. 

7.   It  is  found  that  a  bar  AB,  12 

r— 1  p  ft.  long,  may  be  supported  by  a  force 

• — '  of  18  lbs.  applied  at  the  middle  point. 

IG*  45*  If  it  rests    in    a   horizontal    position 

against  smooth  supports  at  A  and  C, 

4  ft.  apart,  the  support  at  A  being  above  and  that  at  C  below  the 

bar,  what  will  be  the  supporting  forces  ? 

Ans.  27  lbs.  upward  at  C,  9  lbs.  downward  at  A. 

87.  Two  Equal  and  Opposite  Forces.—  If  two  forces  P  and 
Q  are  equal  in  magnitude,  opposite  in  direction,  and  have  different 
lines  of  action,  the  construction  above  given  (Art.  86)  for  finding 
their  resultant  fails,  for  the  reason  that  the  two  lines  AD  and  BD 
(Fig.  43)  become  parallel.  If  it  is  attempted  to  apply  the  general 
rule  for  determining  the  resultant  of  two  opposite  forces,  it  is  found 
that  the  magnitude  of  the  resultant  P — Q  is  zero  ;  while  the  equation 
for  determining  its  line  of  action  becomes  (see  Fig.  43) 

AC/CB=  Q/P=  1, 

which  can  be  true  only  if  AC  and  CB  are  infinite.  Mathematically 
interpreted,  these  results  mean  that  the  resultant  is  a  force  of  magni- 
tude zero,  acting  in  a  line  infinitely  distant  from  the  lines  of  action 
of  the  given  forces. 

Evidently  P  and  Q  in  this  case  form  a  couple  (Art.  53).  A  fur- 
ther discussion  of  couples  will  be  given  later. 

88.  Moment  of  Resultant  of  Two  Forces.  —  Proposition.  — 
The  algebraic  sum  of  the  moments  of  any  two  coplanar  forces  with 
respect  to  a  point  in  their  plane  is  equal  to  the  moment  of  their 
resultant  with  respect  to  that  point. 

In  the  proof  of  this  proposition  two  cases  must  be  considered, 
according  as  the  two  given  forces  are  or  are  not  parallel. 

{a)  Non-parallel  forces. —  For  two  non-parallel  forces,  the  prop- 


COMPOSITION    OF    NON-CONCURRENT    FORCES. 


55 


>R 


■Q 


O 


osition  just  stated  is  a  special  case  of  that  proved  in  Art.  29  for  any- 
localized  vectors.      For  the  resultant  of  two  non-parallel  forces  is  a 
force  equal  to  their  vector  sum,  acting 
in  a  line  through   the  intersection  of 
their  lines  of  action.* 

(fr)  Parallel  forces.  —  Let  P  and  Q 
represent  the  magnitudes  of  any  two 
parallel  forces  and  R  that  of  their  re- 
sultant. Take  any  point  O  (Fig.  46) 
as  origin  of  moments,  and  draw  through 

O  a  line  perpendicular  to  the  forces,  intersecting  their  lines  of  action 
at  A,  B  and  C  respectively.  Let  OA  =  /,  OB  =  q}  OC  =  r. 
From  Art.  86  we  have 

P _BC _r—q 

Q      AC     p-r 

Reducing, 

Pp  +  Qq  =  (P  +  Q)r  =  Rr. 


Fig.  46. 


If  the  forces  P  and  Q  have  opposite  directions,  the  demonstra- 
tion needs  modification.      (See  Fig.  47.)     The  equation  is 


P 

Q 


BC 
AC 


P 


Pp-Qq=(P-Q)r=Rr. 

In  both  cases,  the  reasoning  is  easily  adapted  to  the  case  in  which 
the  origin  of  moments  falls  between  the  lines  of  action  of  the  two 

forces  P  and  Q.      In  all  cases,  the  re- 
sult is   expressed   by  the  proposition 
y(R       Py    Qj^  that  the  moment  of  the  resultant  of  two 

parallel  forces  is  equal  to  the  algebraic 
sum  of  their  moments. 

The  proof  of  the   proposition  for 

Fig.  47.  parallel  forces  may  be  put  in  another 

form,  as  follows :  Referring  to  Art.  86, 

let  any  point  (Figs.  42  and  43)  be  taken  as  origin  of  moments. 

Having  proved  that  the  moment  of  the  resultant  of  two  non-parallel 


0 


*  Any  two  non-parallel  forces  may,  in  fact,  as  already  shown,  be  treated 
as  if  concurrent,  so  that  this  case  reduces  to  that  treated  in  Art.  71. 


56  THEORETICAL    MECHANICS. 

forces  is  equal  to  the  algebraic  sum  of  their  separate  moments,  we 
may  write  the  following  equations: 

mom.  of  R '  =  mom.  of  P  -j-  mom.  of  F' ; 

mom.  of  R"  =  mom.  of  Q  -\-  mom.  of  F" ; 

mom.  of  R  =  mom  of  R'  -j-  mom.  of  R" 

=  mom.  of  P  -f-  mom.  of  Q  -f- 

mom.  ofi**'  -f-  mom.  of  ^ 

=  mom.  of  P  -f-  mom.  of  (2 

(since  the  moments  of  /^'  and  i7"  are  equal  in  magnitude  but  oppo- 
site in  sign). 

89.  Moment  of  a  Couple. — The  proposition  just  proved  is  mean- 
ingless when  applied  to  the  case  of  two  equal  and  opposite  forces, 
since  two  such  forces  are  not  equivalent  to  a  single  resultant  force. 
The  moment  of  a  couple  will  therefore  be  made  a  matter  of  defini- 
tion, as  follows : 

The  moment  of  a  couple  is  the  algebraic  sum  of  the  moments  of 
its  two  forces. 

With  this  definition  it  may  readily  be  shown  that  the  moment  of 
a  couple  has  the  same  value  for  every  origin  in  its  plane,  and  is 
equal  to  the  product  of  the  magnitude  of  either  force  into  the  perpen- 
dicular distance  between  the  lines  of  action. 

§  2.    Couples. 

90.  Equivalent  Couples  in  the  Same  Plane. —  Any  two  couples 
in  the  same  plane  are  equivalent  if  their  moments  are  equal  in  mag- 
nitude and  sign. 

To  prove  this,  consider  first  the  resultant  of  two  couples  whose 
moments  are  equal  in  magnitude  but  opposite  in  sign.  Let  P  de- 
note the  magnitude  of  the  forces  of  one  couple  and  p  the  length  of 
the  arm  or  perpendicular  distance  between  the  lines  of  action  ;  and 
let  Q,  q  denote  the  like  quantities  for  the  second  couple,  these  quan- 
tities being  so  related  that 

Pp  =  Of 

Let  the  lines  of  action  and  directions  of  the  four  forces  be  as  shown 


COMPOSITION   OF    NON-CONCURRENT    FORCES.  57 

in  Fig.  48,  so  that  the  moments  have  opposite  signs.     It  may  be 
shown  that  their  resultant  is  zero. 

Notice  first  that  the  lengths  of  the  sides  of  the  parallelogram 


Fig.  48. 

ABCD  formed  by  the  four  lines  of  action  are  proportional  to  P  and 
Q.     For 

area  ABCD  =  (AB)  ■  q  =  (AD)  -p ; 

and  since  Pp  =  Qqy 

AB/AD  =  Q/P. 

The  resultant  of  the  force  P  acting  along  AD  and  the  force  Q 
acting  along  AB  must  therefore  act  along  the  diagonal  AC  of  the 
parallelogram  ABCD;  and  the  resultant  of  the  force  P  acting  along 
CB  and  the  force  Q  acting  along  CD  must  also  act  along  the  diag- 
onal CA.  But  these  resultants  are  equal,  opposite,  and  collinear; 
hence  their  resultant  is  zero.  The  resultant  of  the  two  couples  is 
therefore  zero. 

Next,  starting  with  the  given  couple  (P,  p)}  let  two  pairs  of  equal 
and  opposite  forces  of  magnitude  Q  be  assumed  to  act  in  the  lines 
AB,  CD  (Fig.  49),  so  taken  that  Qq  =  Pp.  These  four  forces 
form  two  couples ;  one  of  these  counterbalances  the  given  couple  as 
shown  above,  hence  the  other  must  be  equivalent  to  the  given  couple. 
That  is,  the  given  couple  (P,  p)  is  equivalent  to  the  couple  (Q,  q) 
whose  moment  is  equal  to  its  own  in  magnitude  and  sign. 

Since  the  lines  AB  and  DC  may  be  any  two  parallel  lines  inter- 
secting the  lines  of  action  of  the  given  couple,  it  follows  that  any  two 


58 


THEORETICAL    MECHANICS. 


coplanar  couples  whose  forces  are  not  parallel  are  equivalent  if  their 
moments  are  algebraically  equal. 

But  two  couples  whose  forces  are  parallel  and  moments  equal  are 


Fig.  49. 


k2P 


*b 


yftP 


py-. 


equivalent,  because  each  may  be  proved  equivalent  to  a  third  couple  ; 
so  that  the  proposition  holds  for  any  coplanar  couples. 

91.  Couples  in  Parallel  Planes. —  Couples  in  parallel  planes  are 
equivalent  if  their  moments  are  equal  in  magnitude  and  sign. 

Let  AB  (Fig.  50)  be  a  line   perpendicular  to  the  forces  of  a 
couple,  intersecting  their  lines  of  action  in  A  and  B.     Let  CD  be 

any  line  equal  and  parallel  to  AB,  not 
&P  necessarily  in  the  plane  of  the  couple. 

Introduce  at  C  two  equal  and  opposite 
forces,  each  equal  and  parallel  to  P, 
the  force  of  the  given  couple  ;  also 
two  similar  forces  at  D.  The  force  at 
A  and  the  upward  force  at  D  may  be 
replaced  by  an  upward  force  2  P  acting 
Y^>  \fp      at  the  middle  point  of  AD.     The  force 

Fig.  50.  at  B   and   the    downward  force  at    C 

may  be  replaced  by  a  downward  force 
iP  acting  at  the  middle  point  of  CB.  These  two  forces  2P  balance 
each  other,  leaving  two  forces,  —  an  upward  force  at  C  and  a  down- 
ward force  at  D,  each  equal  to  P.  These  form  a  couple  equivalent 
to  the  given  couple. 

Since  the  couple  acting  at  £7  and  D  is  equivalent  to  any  couple 
in  the  same  plane  having  an  equal  moment;  and  since  CD  might  be 


2P 


4P 
D 


COMPOSITION    OF    NON-CONCURRENT    FORCES.  59 

taken  in  any  plane  parallel  to  that  of  the  given  couple,  it  follows 
that  any  two  couples  in  parallel  planes  are  equivalent  if  their  mo- 
ments are  equal. 

92.  Equivalence  of  Couples — General  Result. —  The  results  of 
the  last  two  Articles  may  be  stated  in  one  general  proposition,  as 
follows : 

Any  two  couples  in  the  same  plane  or  in  parallel  planes  are 
equivalent  if  their  moments  are  equal  in  magnitude  and  sign. 

93.  Resultant  of  Coplanar  Couples. —  The  resultant  of  any  num- 
ber of  coplanar  couples  is  a  couple  whose  moment  is  equal  to  the 
algebraic  sum  of  the  moments  of  the  given  couples. 

Choose  any  two  parallel  lines  in  the  plane  of  the  couples,  as  MN, 
M'N'  (Fig.  51),  and  let  each  of  the  given 
couples  be  replaced  by  an  equivalent 
couple  with  forces  acting  in  these  lines. 
Let  the  common  arm  of  these  substituted 
couples  be  /,  and  let  their  forces  be  Px , 
P.2,  P6,  etc.,  their  values  being  either  pos- 
itive or  negative  according  to  the  direc- 
tions of  the  forces,  and  being  such  that 
Pxp,  P2p,  .  .  .  are  algebraically  equal 
to  the  moments  of  the  given  couples.  The 
resultant  of  the  forces  acting  in  MN  is  a 
force  equal  to  their  algebraic  sum  Px  + 

P<i  +  -P-S  +  •  •  •  >  while  the  resultant  of  the  forces  acting  in 
M'N'  has  the  same  magnitude  but  the  opposite  direction.  Hence 
the  whole  system  is  equivalent  to  a  couple  of  moment 

(/>  +  />+/>,+  .  .  .  )p, 

which  is  equal  to 

Pit  +  P*P  +  P*f+     •      •      •     , 
that  is,  to  the  algebraic  sum  of  the  moments  of  the  given  couples. 

The  proposition  obviously  holds  for  couples  lying  in  any  parallel 
planes. 

94.  Resultant  of  Force  and  Couple. —  A  force  and  a  couple 
acting  in  the  same  plane  or  in  parallel  planes  are  equivalent  to  a  sin- 
gle force. 

Let  P  be  the  magnitude  of  the  given  force  and  G  the  moment  of 


60  THEORETICAL    MECHANICS. 

the  given  couple.  Replace  the  couple  by  an  equivalent  couple  in  a 
plane  parallel  to  its  own,  containing  the  given  force  P.  The  magni- 
tude, direction  and  line  of  action  of  one  of  the  forces  of  this  equiva- 
lent couple  may  be  chosen  arbitrarily;  let  them  be  so  chosen  that 
this  force  counterbalances  the  given  force  P.  The  other  force  of  the 
couple  must  be  equal  in  magnitude  and  direction  to  P,  and  its  line  of 
action  must  be  such  that  the  moment  of  the  couple  is  algebraically 
equal  to  G;  its  line  of  action  is  therefore  at  a  distance  G/P  from 
that  of  the  given  force.  This  last  force  is  the  resultant  of  the  given 
force  and  couple. 

Examples. 

i.  A  force  of  50  lbs.  acting  in  any  assumed  line,  and  a  couple 
coplanar  with   it  whose  arm  is    29  ft.  and  whose  forces  have  the 

magnitude  10  lbs.,  have  what  resultant? 
A  g  2.    A  bar  AB,   resting  horizontally 

upon  a  smooth  support  at  B,  is  held  in 

p!     equilibrium  by  two  opposite  horizontal 

forces  applied  at  the  end  A  in  lines  6 

*1G'  52,  ins.  apart.     The  supporting  force  at  B  is 

5  lbs.  and  the  horizontal  forces  are  each 

50  lbs.      If  the  horizontal  forces   cease  to  act  and  the  support  is 

removed,    what  single  force  will  support  the  bar  in   equilibrium  ? 

(Fig-  52.) 

§  3.  Any  Number  of  Coplanar  Forces. 

95.  Resultant  of  Any  Number  of  Forces. —  It  may  now  be 
shown  that  the  resultant  of  any  number  of  coplanar  forces  is  a  single 
force  equal  to  their  vector  sum,  unless  this  sum  is  zero;  in  which 
case  the  resultant  is  a  couple. 

Of  three  or  more  forces,  there  must  be  two  which  do  not  form 
a  couple.  Any  such  pair  may  be  replaced  by  their  resultant,  which 
is  a  single  force  equal  to  their  vector  sum  (Arts.  84,  86),  thus  reduc- 
ing the  number  of  forces  by  one.  This  process  may  be  repeated 
until  the  system  has  been  reduced  to  an  equivalent  system  of  two 
forces.  If  these  two  are  not  equal  and  opposite,  they  are  equivalent 
to  a  single  force  equal  to  their  vector  sum  and  therefore  to  the  vector 
sum  of  the  given  forces.  If  they  are  equal  and  opposite,  they  form  a 
couple  which  is  the  resultant  of  the  given  system;  the  vector  sum  of 
all  the  forces  being  zero. 


COMPOSITION   OF    NON-CONCURRENT    FORCES.  6 1 

96.  Moment  of  Resultant  of  Any  Number  of  Forces. —  In  a 

similar  manner  the  principle  of  moments,  already  proved  for  two 
forces,  may  be  extended  to  any  number  of  coplanar  forces.  That  is, 
it  may  be  shown  that  the  sum  of  the  moments  of  any  number  of  co- 
planar  forces  about  any  origin  in  their  plane  is  equal  to  the  moment 
of  their  resultant  about  that  origin. 

Combining  two  of  the  forces  which  are  not  equal  and  opposite, 
the  sum  of  their  moments  is  (by  Art.  88)  equal  to  the  moment  of 
their  resultant.  Combining  this  resultant  with  another  force  not 
equal  and  opposite  to  it,  the  sum  of  their  moments  is  equal  to  the 
moment  of  their  resultant.  This  process  may  be  continued  until  the 
number  of  forces  is  reduced  to  two.  If  these  two  are  not  equal  and 
opposite,  the  sum  of  their  moments  (equal  to  the  sum  of  the  mo- 
ments of  the  given  forces)  is  equal  to  the  moment  of  the  single  force 
which  is  their  resultant  and  the  resultant  of  the  given  system.  If 
the  two  forces  are  equal  and  opposite,  they  form  a  couple  which  is 
the  resultant  of  the  given  system,  and  whose  moment  is  equal  to  the 
sum  of  the  moments  of  the  given  forces.  Hence  the  proposition  is 
proved. 

97.  Computation  of  Resultant. —  From  the  foregoing  principles, 
the  resultant  of  any  system  of  coplanar  forces  may  be  computed, 
whether  this  resultant  be  a  single  force  or  a  couple. 

(1)  If  the  resultant  is  a  single  force,  its  magnitude  and  direction 
may  be  found  as  if  the  forces  were  concurrent  (Art.  65  or  69) ;  the 
position  of  its  line  of  action  must  be  such  that  its  moment  about  any 
assumed  point  is  equal  to  the  sum  of  the  moments  of  the  given  forces 
about  that  point.  Thus,  if  P  denotes  the  magnitude  of  the  resultant 
force,  and  G  the  sum  of  the  moments  of  the  given  forces,  the  line 
of  action  of  the  resultant  must  be  at  a  distance  G/P  from  the  origin 
of  moments.  This  condition,  together  with  the  requirement  that  the 
sign  of  the  moment  of  the  resultant  is  the  same  as  that  of  the  sum 
of  the  moments  of  the  given  forces,  completely  determines  the  line 
of  action  of  the  resultant. 

(2)  If  the  vector  sum  of  the  given  forces  is  zero,  the  resultant  of 
the  corresponding  concurrent  system  will  be  zero.  The  resultant  of 
the  given  system  is  then  a  couple,  whose  moment  may  be  found  by 
computing  the  sum  of  the  moments  of  the  given  forces  about  any 
point.  The  value  of  this  moment  will  be  the  same,  whatever  point 
be  chosen  as  origin. 


62  THEORETICAL    MECHANICS. 

Let  the  following  examples  be  solved  by  the  direct  application  of 
these  principles. 

Examples. 

[Let  the  magnitude,  direction  and  point  of  application  of  a  force 
be  specified  by  the  following  notation :  x,  y  denote  the  rectangular 
coordinates  of  its  point  of  application ;  a  the  angle  its  direction  makes 
with  the  positive  direction  of  the^r-axis;  Pits  magnitude.] 
i.  Find  the  resultant  of  the  following  forces: 

Px  =  20  lbs.,  Xx  =  2  ft.,  yx  =55  6-ft.,  ax  =  o°; 
P2=so"  x2  =  3  "  72  =  7"  a2=i8o°; 
P3  =  So   "        *,=  — 5"      73  =  7"       «*  =    90°- 

2.  Find  the  resultant  of  a  system  of  parallel  forces  whose  magni- 
tudes and  directions  are  10  lbs.,  24  lbs.,  — 15  lbs.,  3  lbs.,  — 48  lbs.; 
the  successive  distances  between  their  lines  of  action  being  2  ft. ,  3  ft. , 
8  ft,  7  ft. 

3.  Find  the  resultant  of  the  following  forces: 


Px  =  40  lbs. , 

xx  = 

4  ft., 

/i  = 

6  ft., 

ax  =      0°; 

P2=27       " 

*i  = 

2     " 

72    = 

14  - 

a2  =  180°; 

^3=13       " 

x*  = 

—6   M 

73   = 

—8    i( 

a3  =  180°; 

P<=i6     " 

xi  = 

10   " 

74   = 

0    " 

a,  =     90°; 

P5=20       " 

*5   = 

—6   (< 

75   = 

2    " 

«5=       90°; 

P6=36       " 

x6  = 

—  12   4< 

7e  = 

3    " 

a6  =  270°. 

98.  Reduction  to  Force  and  Couple. —  The  general  results  above 
found  regarding  the  resultant  of  any  coplanar  forces  may  be  deduced 

by  a  somewhat  different  line  of  reason- 
ing, as  follows: 

Let  Pu  P2,  .  .  .  ,  represent 
the  magnitudes  of  any  coplanar  forces 
applied  to  a  rigid  body.  Choose  any 
point  O  in  the  plane  of  the  forces  (Fig. 
53),  and  let  the  perpendicular  distances 
from  0  to  the  lines  of  action  of  the  sev- 
J  /  eral   forces    be  plt  p.iy    etc.     Consider 

first  some  one  force  as  Px,     Suppose 
Fig.  53.  two  equal  and  opposite  forces,  parallel 

to  Px  and  equal  to  it  in  magnitude,  to 
be  applied  to  the  body  at  0.  One  of  these  forms,  with  the  given 
force,  a  couple  of  moment  Pxpx.  Hence  the  given  force  Px  is  equiv- 
alent to  a  couple  of  moment  Pxpx  and  a  force  equal  and  parallel  to 
Px  applied  at  O.     If  this  process  is  repeated  for  each  of  the  given 


COMPOSITION   OF    NON-CONCURRENT    FORCES.  63 

forces,  it  is  seen  that  the  given  system  is  equivalent  to  the  following 
two  systems: 

(1)  A  system  of  concurrent  forces,  applied  at  O,  equal  and  par- 
allel to  Px ,  Pt ,     .... 

(2)  A  system  of  couples  whose  moments  are  Pxpx,  P2p2,    .    .    .    . 
The  resultant  of  this  system  of  concurrent  forces  may  be  found  as 

in  Art.  69,  while  the  couples  may  be  combined  as  in  Art.  93.     Hence 
the  following  proposition: 

Any  system  of  coplanar  forces  is  equivalent  to  a  single  force  equal 
to  their  vector  sum,  applied  at  any  chosen  point;  together  with  a 
couple  whose  moment  is  the  algebraic  sum  of  the  moments  of  the 
given  forces  with  respect  to  that  point. 

99.  Computation  of  the  Force  and  Couple. —  Let  the  point  O  at 
which  the  concurrent  forces  are  assumed  to  act  be  taken  as  origin  of 
coordinates,  and  let  the  angles  made  with  the  x-  and  jj/-axes  by  any 
force  P  be  denoted  by  a,  /3,  with  proper  suffix. 

Let  the  force  and  couple  to  which  the  system  is  reduced  be  speci- 
fied by  the  following  notation: 

R  =  magnitude  of  force; 
a  =  angle  between  R  and  ^r-axis; 
b  =  angle  between  R  and  j/-axis; 
Xy  Y  =  axial  components  of  R  ; 
G  =  moment  of  couple. 
Then 

X  =  R  cos  a  —  Px  cos  ax  +  P*  cos  a2  +    •     •     • 

Y  =  R  cos  b  =  Px  cos  &  +  P2  cos  &  +    .     .     . 


(0 


R  =  VX2  +..y»; (2) 

cos  a  =  XI R  ;  cos  b  =  Y/R (3) 

Also,  by  Art.  98, 

ff«/VA  +  /^%+    •■•.'•■■>■■.        •       •       ■'  <4) 

The  values  of  R,  a  and  b  completely  determine  the  required 
force ;  and  the  value  of  G  determines  the  moment  of  the  couple. 

It  is  seen  that  the  magnitude  and  direction  of  the  force  R  are  the 
same  wherever  the  point  O  be  taken;  while  in  general  the  value  of  G 
depends  upon  the  position  of  the  point  at  which  the  concurrent  forces 
are  assumed  to  act. 


64  THEORETICAL    MECHANICS. 

Let  the  following  examples  be  solved  by  this  process. 

Examples. 

1.  Find  a  force  and  couple  equivalent  to  the  following  forces: 
Px  =  70  lbs.,     xx  =  4  ft.,     yx  =        8.  ft.,     al  =    6o°; 
P2  =  4o    "       x,  =  6    '•      yt  =        6   "      «,=    450; 
P.A=25    "       x3  =  o   "      y3  —  — 10   "      a3=i2o\ 

2.  Reduce  each  of  the  systems  of  forces  given  in  examples  1  and 
3,  Art.  97,  to  a  couple  and  a  force  applied  at  the  origin  of  coordinates. 

100.  Resultant  Force  or  Resultant  Couple. —  The  force  and 
couple  to  which  any  system  may  be  reduced  by  the  above  process 
may  in  general  be  combined  into  a  simpler  resultant.  Since  in  par- 
ticular cases  one  or  both  the  quantities  R  and  G  determined  by  the 
above  process  may  reduce  to  zero,  the  following  four  cases  must  be 
considered : 

(1)  Suppose  neither  R  nor  G  is  equal  to  zero.  In  this  case  the 
force  and  couple  may  be  combined  into  a  single  force  (Art.  94)  hav- 
ing the  same  magnitude  and  direction  as  R;  its  line  of  action  being 
distant  G/R  from  O. 

(2)  Suppose  G  =  o.  In  this  case  the  whole  system  reduces  to 
the  single  force  R,  its  line  of  action  passing  through  the  assumed 
point  O.  [This  case  always  results  if  the  given  system  is  equivalent 
to  a  single  resultant  force,  and  if  the  point  0  happens  to  be  chosen 
upon  its  line  of  action.  ] 

(3)  Let  R  =  o.  In  this  case  the  resultant  of  the  given  system 
is  a  couple  whose  moment  is  G. 

(4)  Let  R  =  o,  G  =  o.  In  this  case  the  given  system  is  equiv- 
alent to  no  force.  In  other  words  the  given  forces  exactly  balance 
each  other  and  the  system  is  in  equilibrium.* 

Examples. 

In  each  of  the  examples  of  Art.  99,  determine  completely  the 
resultant  force  or  resultant  couple. 

*  In  the  last  three  Articles  we  have  reached  by  means  of  algebraic  analysis 
the  same  results  deduced  previously  by  geometrical  reasoning.  The  student 
will  find  it  profitable  to  become  familiar  with  both  methods  of  treatment. 


CHAPTER  V. 

EQUILIBRIUM    OF    COPLANAR    FORCES. 

§  i.    General  Principles  of  Equilibrium. 

101.  Meaning  of  Equilibrium. —  The  word  equilibrium  is  used 
with  reference  both  to  forces  and  to  bodies.     (See  Art.  57.) 

Any  number  of  forces  form  a  system  in  equilibrium  if  their  com- 
bined action  produces  no  effect  upon  the  motion  of  the  body  to 
which  they  are  applied. 

A  rigid  body  is  in  equilibrium  if  all  external  forces  acting  upon  it 
form  a  system  in  equilibrium. 

102.  General  Condition  of  Equilibrium. —  It  follows  from  the 
definition  of  equilibrium  that  the  necessary  and  sufficient  condition  of 
equilibrium  for  any  system  of  forces  is  that  their  resultant  is  zero. 
This  general  condition  implies  subordinate  conditions  which  are  now 
to  be  considered. 

These  subordinate  conditions  may  be  deduced  in  two  ways : 
First,  by  the  direct  application  of  the  principles  regarding  the  re- 
sultant deduced  in  Arts.  95  and  96  ;  or  second,  by  applying  the  re- 
sults of  the  algebraic  discussion  of  Arts.  98-100.  It  will  be  useful  to 
consider  both  methods. 

103.  Equations  of  Equilibrium. —  It  has  been  shown  (Art.  95) 
that  the  resultant  is  either  a  force  or  a  couple.  If  a  force,  it  is  equal 
to  the  vector  sum  of  the  given  forces,  and  the  resolved  part  of  the 
resultant  in  any  chosen  direction  must  equal  the  algebraic  sum  of 
the  resolved  parts  of  the  given  forces  in  that  direction.  If  the  re- 
sultant is  a  couple,  the  vector  sum  of  the  forces  is  zero,  and  the 
algebraic  sum  of  their  resolved  parts  in  any  direction  is  zero. 
In  either  case,  the  sum  of  the  moments  of  the  given  forces  about 
any  origin  is  equal  to  the  moment  of  the  resultant  (Art.  96).  Hence, 
if  the  resultant  is  zero,  the  following  conditions  must  be  satisfied: 

(1)  The  sum  of  the  resolved  parts  of  the  given  forces  in  any 
direction  is  zero. 

(2)  The  sum  of  the  moments  of  the  given  forces  about  any  point 
is  zero. 


66  THEORETICAL    MECHANICS. 

From  these  two  principles,  an  infinite  number  of  equations  may 
be  written  ;  since  the  forces  may  be  resolved  in  any  direction,  and 
any  point  may  be  taken  as  origin  of  moments.  It  may  be  shown, 
however,  that  only  three  of  these  equations  can  be  independent. 

104.  Three  Independent  Equations.  —  In  order  to  determine 
how  many  of  the  possible  equations  of  equilibrium  can  be  indepen- 
dent, consider  how  much  is  implied  by  any  one  of  them. 

(a)  If  the  sum  of  the  resolved  parts  of  the  forces  is  zero  for  any 
given  direction  of  resolution,  the  resultant  force,  if  one  exists,  must 
be  perpendicular  to  that  direction.  There  may,  however,  be  a  re- 
sultant couple. 

(d)  If  the  sum  of  the  moments  is  zero  for  any  chosen  origin, 
there  can  be  no  resultant  couple,  since  the  moment  of  a  couple  is 
not  zero  for  any  origin  ;  if  there  is  a  resultant  force,  its  line  of  action 
must  pass  through  the  origin. 

These  principles  lead  immediately  to  the  following  propositions: 

(1)  There  will  be  equilibrium  if  the  sum  of  the  resolved  parts  is 
zero  for  each  of  two  directions  of  resolution,  and  the  sum  of  the  mo- 
ments is  zero  for  one  origin. 

For,  by  principle  (a),  there  can  be  no  resultant  force,  since  such 
a  force  would  need  to  be  perpendicular  to  both  directions  of  resolu- 
tion; and  by  principle  (J?)  there  can  be  no  resultant  couple,  since  the 
moment  of  a  couple  cannot  be  zero. 

(2)  There  will  be  equilibrium  if  the  sum  of  the  moments  is  zero 
for  each  of  the  two  origins,  and  the  sum  of  the  resolved  parts  is  zero 
in  any  one  direction,  not  perpendicular  to  the  line  joining  the  two 
origins. 

For,  by  principle  (b),  there  can  be  no  resultant  couple,  and  the 
resultant  force  (if  one  exists)  must  act  in  a  line  through  the  two 
origins  of  moments  ;  while  by  principle  (a),  the  resultant  force  (if 
there  is  one)  must  be  perpendicular  to  the  direction  of  resolution. 

(3)  There  will  be  equilibrium  if  the  sum  of  the  moments  is  zero 
for  each  of  three  origins  not  in  the  same  straight  fine. 

For,  by  principle  (&),  the  resultant  force  (if  one  exists)  must  act 
in  a  line  containing  the  three  origins;  and  there  can  be  no  resultant 
couple,  since  the  moment  of  a  couple  cannot  be  zero. 

Since,  therefore,  three  equations  can  be  written,  which,  if  satisfied, 
insure  that  the  resultant  is  zero,  it  follows  that  all  possible  equations 
obtained  in  accordance  with  the  principles  of  Art.  103  must  be  true 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES.  67 

if  those  three  are  true.  Hence  three,  and  only  three,  of  the  equa- 
tions of  equilibrium  can  be  independent. 

105.  Conditions  of  Equilibrium  Deduced  Algebraically. —  It  was 

shown  in  Art.  98  that  any  system  of  coplanar  forces  can  be  reduced 
to  a  force  and  a  couple;  the  force  being  applied  at  any  chosen  point 
and  being  equal  to  the  vector  sum  of  the  given  forces,  while  the  mo- 
ment of  the  couple  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  given  forces  about  the  chosen  point. 

Using  the  notation  of  Art.  99,  the  equations  for  computing  the 
force  and  couple  are 

X  =  R  cos  a  =  Px  cos  a  l  -f-  Py  cos  a  ,  -f     .     .     .      ; 

Y  =  R  sin  <z  =  Px  cos  #rfp  P2  cos  /32  +     .     .     .     ; 
Rl  =  X'1  +  Y2- 

G  =  Pxpx  +  />/,+     .... 

The  analysis  of  the  four  possible  cases,  given  in  Art.  100,  shows 
that  there  will  be  equilibrium  if  R  and  G  are  both  zero,  but  not 
otherwise. 

Now  the  condition  R  =  o  requires  that  X  =  o  and  Y  =  o,  un- 
less either  X'2  or  Y2  is  negative,  that  is,  unless  X  or  Y  is  imaginary. 
Butif/^,/^,     .     .     .     are  real  forces,  X and  Y must  be  real. 

For  equilibrium,  therefore,  it  is  necessary  that  the  three  equa- 
tions 

Px  cos  ax  -f"  P%  cos  a2  "T"     •      •      •     —  °>    •  •     (0 

/\  cos  £/+/>,  cos  £,+     .     .     .     =0,    .         .     (2) 

<PiA.+  ^.A+  •  •  •  =0,  .  .  (3) 
shall  be  satisfied.  And  conversely,  if  these  three  equations  are  satis- 
fied, the  system  is  in  equilibrium. 

Since  the  origin  and  axes  may  be  chosen  at  pleasure,  an  infinite 
number  of  sets  of  equations  similar  to  (1),  (2)  and  (3)  may  be  writ- 
ten.    If  one  of  these  sets  is  satisfied,  all  others  must  be. 

106.  Parallel  Forces. —  The  general  conditions  of  equilibrium 
deduced  above  apply  to  any  system  of  coplanar  forces.  In  case  all 
the  forces  are  parallel,  however,  only  two  independent  equations  of 
equilibrium  can  be  written.  For,  from  principles  {a)  and  (b),  Art. 
104,  it  is  evident  that  a  system  of  parallel  forces  will  be  in  equi- 
librium in  either  of  the  following  cases  : 


68  THEORETICAL   MECHANICS. 

(i)  If  the  sum  of  the  moments  is  zero  for  each  of  two  origins  not 
lying  on  a  line  parallel  to  the  forces. 

(2)  If  the  sum  of  the  moments  is  zero  for  one  origin,  and  the  sum 
of  the  resolved  parts  is  zero  for  any  direction  not  perpendicular  to 
the  forces.* 

107.  Special  Condition  of  Equilibrium. —  If  any  system  of 
forces  in  equilibrium  be  divided  into  two  sets,  the  resultants  of  these 
sets  must  be  equal  and  opposite  and  have  the  same  line  of  action. 

If  three  forces  are  in  equilibrium,  their  lines  of  action  must  meet 
in  a  point,  or  be  parallel.  For  the  resultant  of  any  two  must  be 
equal  and  opposite  to  the  third,  and  have  the  same  line  of  action. 

This  principle  is  often  found  useful  in  the  solution  of  problems  in 
equilibrium. 

§  2.  Application  of  Principles  of  Equilibrium. 

108.  General  Method  of  Solving  Problems  in  Equilibrium. — 

The  solution  of  problems  in  equilibrium  is  the  most  important  of  the 
practical  applications  of  the  principles  of  Statics.  The  problems  to 
be  solved  are  of  the  following  kind : 

A  body  is  in  equilibrium  under  the  action  of  any  number  of 
forces,  some  of  which  are  partly  or  wholly  unknown  ;  it  is  required 
to  determine  these  unknown  forces  completely. 

The  general  method  of  solving  such  a  problem  is  to  write  three 
independent  equations  of  equilibrium,  introducing  as  many  unknown 
quantities  as  necessary  to  represent  completely  all  the  forces  acting 
on  the  body.  The  unknown  quantities  must  then  be  determined  by 
solving  the  equations. 

If  the  number  of  unknown  quantities  is  greater  than  three  for 
non-parallel  and  non-concurrent  forces  (or  greater  than  two  for  par- 
allel or  concurrent!  forces),  the  equations  of  equilibrium  are  not 
sufficient  for  their  determination.  The  problem  is  then  indetermi- 
nate, unless  additional  equations  can  be  written  from  the  geometrical 
relations. 

*  Placing  the  sum  of  the  resolved  parts  in  any  direction  equal  to  zero,  the 
same  equation  results,  whatever  the  direction  of  resolution;  all  such  equations 
reduce  to  the  form,  "algebraic  sum  of  forces  =  0." 

t  Forces  applied  to  the  same  rigid  body  may  be  regarded  as  concurrent 
if  their  lines  of  action  intersect  in  a  single  point,  even  if  they  are  not  actually 
applied  at  that  point.    (Art  82.) 


EQUILIBRIUM   OF    NON-CONCURRENT    FORCES.  69 

The  equations  of  equilibrium  may  often  be  simplified  by  a  judi- 
cious selection  of  the  directions  of  resolution  and  of  the  origins  of 
moments.  Thus,  an  equation  free  from  any  one  force  may  be  ob- 
tained by  resolving  in  a  direction  perpendicular  to  that  force,  or 
by  taking  moments  about  an  origin  lying  on  its  line  of  action.  If 
moments  are  taken  about  the  point  of  intersection  of  the  lines  of 
action  of  two  forces,  neither  of  these  forces  will  enter  the  resulting 
equation. 

In  solving  problems  in  the  equilibrium  of  rigid  bodies,  the  stu- 
dent may  be  aided  by  the  following  outline  of  the  method  of  pro- 
cedure: 

(1)  Specify  the  body  to  which  the  conditions  of  equilibrium  are 
to  be  applied. 

(2)  Enumerate  all  forces  acting  upon  this  body,  specifying  the 
magnitude  and  direction  of  each  so  far  as  known. 

(3)  Write  three  independent  equations  of  equilibrium  (two,  if  the 
forces  are  known  to  be  concurrent  or  parallel),  introducing  as  many 
unknown  quantities  as  necessary.  In  writing  these  equations,  notice 
what  directions  of  resolution,  or  what  origins  of  moments,  give  the 
simplest  equations. 

(4)  Notice  whether  the  number  of  equations  is  as  great  as  the 
number  of  unknown  quantities.     If  not, 

(5)  Notice  whether  any  geometrical  equations  can  be  written. 
Write  as  many  of  these  as  possible. 

(6)  Notice  whether  the  number  of  statical  and  geometrical  equa- 
tions together  is  sufficient  for  the  determination  of  the  unknown 
quantities. 

(7)  If  the  problem  is  found  to  be  determinate,  solve  the  equa- 
tions algebraically,  thus  determining  the  unknown  quantities. 

After  some  experience  the  student  will  be  able  in  most  cases  to 
select  readily  the  best  methods  of  writing  the  equations  of  equi- 
librium, and  will  often  be  able  to  solve  problems  by  short  methods. 
For  example,  the  principle  stated  in  Art.  107  will  often  be  found 
useful.  The  beginner  will,  however,  usually  find  it  useful  to  analyze 
the  problem  completely  in  a  manner  similar  to  that  outlined  above. 

In  all  cases  the  three  equations  of  equilibrium  should  be  written, 
even  if  the  complete  solution  of  the  problem  is  found  to  be  difficult 
or  impossible.  When  these  have  been  correctly  written,  the  problem 
is  solved  so  far  as  the  application  of  the  principles  of  Mechanics  is 


yo 


THEORETICAL    MECHANICS. 


concerned.  The  remainder  of  the  process  is  a  matter  of  algebraic 
manipulation. 

This  general  method  will  now  be  illustrated. 

109.  Problems  in  Equilibrium  of  a  Rigid  Body.—  I.  A  rigid 
bar  AB  (Fig.  54)  rests  with  the  end  A  against  a  horizontal  floor  and 
a  vertical  wall,  the  other  end  being  supported  by  a  string  BC}  and  a 
weight  of  20  lbs.  being  suspended  by  a  string  from  the  point  B. 

Let  AB  and  BC  each  make  an  angle 
of  300  with  the  horizontal,  and  let 
the  weight  of  the  bar  be  5  lbs.,  its 
point  of  application  being  taken  as 
the  middle  point  of  AB.  Determine 
all  forces  acting  upon  the  bar. 

Solution. —  Following  the  method 
outlined  above,  we  proceed  as  follows : 

(1)  The  bar  AB  is  the  body  to 
which  the  conditions  of  equilibrium 
will  be  applied. 

(2)  The  forces  acting  on  the  bar 
are  five  in  number:  a  force  of  20  lbs. 

vertically  downward  at  B,  due  to  the  suspended  weight ;  a  down- 
ward force  of  5  lbs,  at  D,  the  center  of  gravity  of  the  bar ;  a  force 
applied  at  B  in  the  direction  BC,  due  to  the  supporting  string;  a 
pressure  at  A  exerted  by  the  floor ;  and  a  pressure  at  A  due  to  the 
wall.  The  last  two  forces  may  be  replaced  by  their  resultant,  which 
is  a  force  in  some  unknown  direction,  but  applied  at  A.  Let  H  and 
V  represent  the  horizontal  and  vertical  components  of  this  force; 
and  let  Q  denote  the  force  applied  at  B  along  the  line  BC. 

(3)  We  will  choose  as  the  equations  of  equilibrium  (a)  a  moment 
equation  with  origin  A  ;  (J?)  a  moment  equation  with  origin  C;  and 
(c)  a  resolution  equation,  resolving  along  A  C.  Denoting  the  length 
of  the  bar  by  2a*  and  noticing  that  the  triangle  ABC  is  equilateral, 
the  three  equations  are  as  follows: 

Q  •  2a  sin  6o°  —  20  •  2a  cos  30°  —  5  •  a  cos  30°  =  o ;    (a) 

H  •  2a  —  20  •  2a  cos  300  -*-  5  •  a  cos  300  —  b ;  (b) 

V  -jr  Q  cos  6o°  —  5  —  20  =  o (r) 


Fig.  54. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


71 


(4,  5,  6)  These  equations  are  just  sufficient  to  determine  the 
three  unknown  quantities.  No  geometrical  equations  are  needed; 
in  fact  the  geometrical  relations  have  been  wholly  taken  account  of  in 
writing  the  three  equations  of  equilibrium. 

(7)  From  equation  (a),  Q  =  22.5  lbs.  From  ($),  H  =  19.48 
lbs.     From  (V),   F=  13.7 5  lbs. 

The  resultant  of  the  forces  exerted  by  the  floor  and  wall  at  A  is 
equal  to  the  resultant  of  H  and  V.  Its  magnitude  is  V Hl  -\-  V% 
=  23.85  lbs. ;  its  direction  is  inclined  to  the  horizontal  at  an  angle  0 
such  that  tan  6  =  V\H  =  0.706,  hence  6  =  350  13'. 

The  components  of  P  which  are  exerted  by  the  floor  and  wall 
respectively  cannot  be  determined  without  further  data.  If  the  floor 
and  wall  are  supposed  perfectly  smooth,  the  pressure  exerted  by  each 
is  normal  to  itself;  hence  in  this  case  Fis  the  force  exerted  by  the 
floor  and  H  the  force  exerted  by  the  wall. 

Geometrical  solution. —  The  problem 
may  be  solved  by  aid  of  the  principle  of 
Art.  107,  as  follows  : 

Let  the  two  vertical  forces  of  5  lbs. 
and  20  lbs.  be  replaced  by  their  resultant, 
R,  a  downward  force  of  25  lbs.  whose  line 
of  action  divides  BD  into  segments  of 
lengths  a/5  and  40/5  (Art.  86).  Let  this 
line  of  action  intersect  BC  at  E  (Fig.  55); 
then  the  force  P  acting  upon  the  bar  at  A 
must  act  through  the  point  E  (Art.  107), 
and  its  line  of  action  is  therefore  AE.  The 
three  forces  P,  Q  and  R  are  in  equi- 
librium, hence  their  vector  sum  is  zero. 
Representing  R   by  the  vector  EG,  and 

drawing  GH  parallel  to  EC  and  FH  parallel  to  EA,  the  vectors 
GH  and  HF  must  represent  Q  and  P  in  magnitude  and  direction. 
The  magnitudes  P  and  Q  may  be  found  from  the  geometrical  rela- 
tions, thus :  The  triangle  ABC  being  equilateral,  we  have  (Fig.  55) 

CA  =BC=2a; 

BE  =  BD/s  =  a/s  J 

EC  =  2a  —  a  1 5  =  ga/ 5. 


Fig.  55. 


72 


THEORETICAL    MECHANICS. 


or 


Also 


Since  CAE  and  FGH  are  similar  triangles,  we  have 

GHJFG  =  EC/CA, 

QIR  =  (ga/s) ■■*  (2*)=  gfio; 

2  =  0.9  7?  =  22.5  lbs. 

P2  =  Q2  +  i?2  —  2(2^  cos  6o° 

=  (22. 5)2  +  (25)'  —  2  X  22.5  X  25  X  0.5 

=  568.75; 
P=  23.85  lbs. 

To  find  the  direction  of  P,  we  have  {6  being,  as  above,  the  angle 
between  P  and  the  horizontal) 

sin  6  =  cos  (HFG)  =  (P2  +  R2  —  Q2)/2PR  ==  0.5767  ; 


e 


35"  13 


II.  A  bar  AB  (Fig.  56)  of  mass  PFlbs.,  whose  center  of  gravity 

is  at  any  point,  rests  with  the  end  A  on  a  smooth  horizontal  plane 

and  the  end  B  against  a  smooth  vertical  plane.     At  A  is  attached  a 

string  which  passes  over  a  smooth  peg  at 

C  and  sustains  a  body  of  P  lbs.     Find  the 

position  of  equilibrium  and   the  pressures 

exerted  upon  the  bar  by  the  smooth  planes. 

Solution. — The  center  of  gravity  of  the 

bar  being   at    G,  let  AG  —  a,  BG  =  b 

(known    quantities),   and  let  angle  BAC 

=  6    (unknown).       Following   the   above 

outline  we  have: 

(1)  The  conditions  of  equilibrium  are 
to  be  applied  to  the  bar  AB. 

(2)  The  forces  acting  upon  the  bar  are 
four :  its  weight  W  lbs. ,  acting  vertically 

downward  at  G  ;  the  horizontal  pull  of  P  lbs.  exerted  by  the  string 
at  A  ;  the  pressure  of  the  plane  at  A,  directed  vertically  upward,  its 
unknown  magnitude  being  called  R  ;  the  pressure  of  the  plane  at  B, 
directed  horizontally,  its  unknown  magnitude  being  called  6". 

(3)  For  the  equations  of  equilibrium,  let  three  moment  equations 
be  written  as  follows: 


pQ 


Fig.  56. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES.  73 

With  origin  at  A, 

S  (a  +  b)  sin  0  —  Wa  cos  0  =  o.         .         .     (i) 
With  origin  at  B, 

P  {a  +  b)  sin  0  —  R  {a  +  b)  cos  d  +  Wb  cos  0  =o.        (2) 
With  origin  at  C, 

S  (a  +  b)smd  —  R(a  +  b)  cos  0  +  Wb  cos  0  =  0.        (3) 

(4)  The  unknown  quantities  are  R,  S  and  6,  three  in  number; 
hence  the  three  equations  are  sufficient. 

(7)  By  inspection  of  equations  (2)  and  (3)  it  is  seen  that 

S  =  P. 

From  equation  (1), 

tan  6  =  Wa/S(a  -f  £)  =  Wa  j  P(a  +  *). 
From  equation  (2), 

7e  =  Ptan  6  +   Hft/fc  +  *)*=  W. 

[If  two  of  the  equations  of  equilibrium  had  been  those  obtained 
by  resolving  forces  horizontally  and  vertically,  the  relations  R  =  W 
and  5  =  P  would  have  appeared  at  once.  The  above  method  was 
chosen  to  illustrate  the  sufficiency  of  three  moment  equations.] 

Geometrical  solution. — Representing  by  Q  the  resultant  of  P 
and  Wy  the  line  of  action  of  Q  must  pass  through  E  (Fig.  56). 
Again,  the  line  of  action  of  the  resultant  of  R  and  5  must  pass 
through  D.  These  resultants  must  be  equal  and  opposite  and  act 
in  the  same  line  DE.     We  have  therefore 

DAIAE=  W/P; 

or  since  AE  =  a  cos  6  and  DA  =  (a  -\-  b)  sin  0, 

tan  d  =  Wa/P(a  +  b). 

Also,  since  the  resultant  of  R  and  S  is  equal  and  opposite  to  Q, 
and  since  6"  is  parallel  to  P  and  R  parallel  to  W,  the  triangle  of 
forces  shows  that  R  =  Wand  S  =  P.  The  triangle  of  forces  is  not 
shown,  but  should  be  drawn  by  the  student. 

III.  A  straight  bar  AB  (Fig.  57),  whose  center  of  gravity  is  at 
its  middle  point,  rests  with  one  end  B  against  a  smooth  vertical  wall 


74 


THEORETICAL    MECHANICS. 


and  the  other  end  upon  a  smooth  horizontal  floor.     A  string  is 
attached  to  the  bar  at  A  and  is  fastened  to  the  wall  at   C.     The 
lengths  AB,  AC  and  AD  being  given,  dis- 
cuss the  possibility  of  equilibrium. 

Solution.—  Let  AB  =  /;  AC  =  a;  AD 
=  b  ;  angle  BAD  =  a  ;  angle  CAD  =  /3. 
Following  the  outline  given  in  Art.  108  we 
proceed  as  follows: 

(i)  The  body  whose  equilibrium  is  to 
be  considered  is  the  bar  AB. 

(2)  The  forces  acting  upon  the  bar  are 
four :  its  weight  W  lbs.  acting  vertically 
downward  at  its  middle  point ;  the  pressure 
(P  lbs. ,  unknown)  exerted  by  the  wall  at  B, 
its  direction  being  horizontal  since  the  wall  is  smooth;  the  pressure 
(<2  lbs.,  unknown)  exerted  by  the  floor  at  As  its  direction  being 
vertically  upward ;  the  force  (R  lbs. ,  unknown)  exerted  by  the 
string  at  A,  its  line  of  action  being  AC. 

(3)  Resolving  forces  horizontally  and  vertically,  and  taking  mo- 
ments about  A}  we  obtain  the  following  three  equations: 

— P-\-  Rcos  /3  =  o]  .         .         .     (1) 

Q  +  R  sin  j3  —  W  =  o ;  .         .     (2) 


Fig.  57. 


P/sin  a 


Wl  cos  a  =  o. 


(3) 


(4)  The  unknown  quantities  are  only  three,  P,  Q  and  Ry  equal 
in  number  to  the  equations,  aside  from  the  angles  a  and  /3,  whose 
values  are  easily  expressed  in  terms  of  /,  a  and  b. 

(5)  The  geometrical  relations  are  merely  the  values  of  cos  a  and 
cos  /3  in  terms  of  the  known  lengths.     They  are 

cos  a  =  b//;     cos  /S  =  bja. 

(7)  To  solve  the  equations,  proceed  as  follows :  From  equation  (3), 

Wb 


V2  W  cotan  a  = 


2V/2  —  b- 


From  equation  (1), 

R  = 


cos  y8 


Pa 
b 


Wa 


2vr 


EQUILIBRIUM    OF    NON-CONCURRENT   FORCES. 

From  equation  (2), 

Q  =  W  —  R  sin  /3  =  W 


75 


Va2  —  b2 


2V i*.  —  b 

The  unknown  forces  are  thus  completely  determined. 

A  closer  analysis  shows,  however,  that  in  certain  cases  the  solu- 
tion fails.  The  force  P  cannot  act  toward  the  right,  and  the  force  Q 
cannot  act  downward.  If,  therefore,  either  of  the  above  values  of  P 
and  Q  becomes  negative,  the  solution  fails.  In  such  a  case  equi- 
librium is  impossible.  The  value  found  for  P  will  always  be  positive; 
but  Q  will  be  negative  if 

a2  -  b2  >  4(l2  -  b2), 

that  is  if  a2  >  4/*  —  3b2. 

Numerical  case. —  Suppose  AB  =  /  =  8  ft. ;  AC =  a  =  10  ft. ; 
AD  —  b  =  6  ft.     Then 


P  = 


R 


Wb 


e=  w 


1  — 


2v 72- 

Wa 

b2 

2VI2- 

b2 

Va2- 

-b2 

2|/28 


M/=  0.567  W\ 


21/7 


^='■0.945  W\ 


y-p 


a  ^ [i  —  -1/7 1  =  0.244  rr. 


This  value  of  Q  being  positive,  equilibrium  is  possible, 
be  foreseen  from  the  condition  above  deduced ;  for 


This  might 


00 


hence 


4/2-3^=  148; 


If,  however,  we  take  b  =  7.5  ft.,  the  other  data  remaining 
unchanged,  the  value  of  Q  is  negative,  and  the  solution  fails. 

Geometrical  analysis. — The  resultant  of  P  and  W  acts  through 
the  intersection  of  their  lines  of  action  (F,  Fig.  57) ;  hence  for  equi- 
librium the  resultant  of  Q  and  R  must  act  through  F.  Since  Q  must 
act  upward,  the  resultant  of  Q  and  R  acts  in  some  line  between  A  C 
and  the  vertical;  hence  this  line  cannot  pass  through  Sunless  .Flies 


76  THEORETICAL    MECHANICS. 

above  A  C.     In  the  limiting  case  in  which  A  C  passes  through  F, 
we  have,  from  the  geometrical  relations, 

a2  =  4/2  —  3&\ 

Hence  if  a*  >  \P  —  3#2,  equilibrium  is  impossible.     This  agrees 
with  the  result  reached  above. 


Examples. 

i.  A  bar  of  mass  W  lbs.  and  length  /  rests  with  one  end  upon  a 
smooth  horizontal  floor  and  the  other  against  a  smooth  vertical  wall, 
being  held  in  equilibrium  by  a  horizontal  string  attached  at  a  point 
distant  b  from  the  upper  end.  The  center  of  gravity  is  at  a  distance 
a  from  the  upper  end,  and  the  inclination  to  the  horizontal  is  0. 
Determine  completely  all  forces  acting  upon  the  bar. 

Ans.  Tension  in  string  =  IV  (/ —  a)/ b  tan  0. 

2.  In  Ex.  i,  let  W  =  20  lbs.,  0  =  40°,  a  ==  0.6/,  b  =  0.3/;  de- 
termine all  forces  completely. 

Ans.  Reaction  at  lower  end  =  20  lbs. ;  at  upper  end,  31.79  lbs. ; 
tension  =  31.79  lbs. 

3.  A  bar  of  mass  1 5  lbs. ,  whose  center  of  gravity  is  at  its  middle 
point,  rests  with  its  ends  upon  two  smooth  planes  inclined  to  the 
horizontal  at  angles  of  36°  and  450  respectively.  Determine  the  in- 
clination of  the  bar  to  the  horizontal  when  in  equilibrium;  also  the 
pressures  exerted  upon  it  by  the  supporting  planes. 

Ans.  io°  39';  8.93  lbs. ;   10.74  lbs. 

4.  A  bar  of  known  mass  and  length  being  supported  as  in  the 
preceding  example,  let  its  center  of  gravity  be  at  any  point  of  its 
length,  and  let  a  and  /3  be  the  angles  made  by  the  planes  with  the 
horizontal.  Determine  the  position  of  equilibrium  and  the  support- 
ing pressures. 

5.  A  person  weighing  160  lbs.,  standing  upon  a  scale  platform, 
pulls  upon  a  suspended  rope  in  a  direction  inclined  io°  to  the  ver- 
tical. The  scale  beam  shows  his  apparent  weight  to  be  140  lbs. 
With  what  force  does  he  pull  the  rope  ?  Ans.  20. 3  lbs. 

6.  A  bar  AB,  16  ft.  long,  is  supported  in  a  horizontal  position  by 
a  smooth  hinge  at  A  and  a  smooth  plane  at  B,  the  inclination  of  the 
plane  to  the  horizontal  being  20°.  Weights  of  20  lbs.,  18  lbs.  and 
28  lbs.  are  suspended  from  the  body  by  cords  attached  at  points 
whose  distances  from  A  are  6  ft.,  10  ft.  and  15  ft.  Determine  the 
supporting  forces  exerted  by  the  plane  and  hinge.  (The  pressure 
exerted  by  the  hinge  may  have  any  direction,  but  always  acts  through 
the  center  of  the  hinge.     See  Art.  42.) 

Ans.  Pressure  at  B  =  47.89  lbs. ;  pressure  at  A  =  26.63  lbs. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES.  77 

7.  A  bar  whose  mass,  center  of  gravity  and  length  are  known,  is 
supported  by  strings  attached  to  the  ends.  In  order  that  the  bar 
may  rest  in  a  horizontal  position,  what  condition  must  be  satisfied  by 
the  directions  of  the  strings  ?     Solve  geometrically. 

8.  A  bar  AB  whose  center  of  gravity  is  at  a  distance  from  the 
end  A  equal  to  one-third  the  length,  is  supported  as  in  the  preced- 
ing example.  If  the  string  at  A  is  inclined  30°  to  the  vertical,  what 
must  be  the  inclination  of  the  string  at  B  ?  Determine  the  tensions 
in  the  strings,  the  mass  of  the  bar  being  W  lbs. 

Ans.  490  7';  0.77  W ';  0.509  W. 

9.  A  bar  AB,  whose  center  of  gravity  is  at  its  middle  point,  and 
whose  mass  is  1 2  lbs. ,  is  supported  in  a  horizontal  position  by  strings 
attached  to  the  ends,  and  sustains  loads  of  16  lbs.  and  20  lbs.  at  A 
and  B  respectively  If  the  string  at  A  is  inclined  450  to  the  hori- 
zontal, what  is  the  direction  of  the  string  at  B  ?  What  tensions  are 
sustained  by  the  strings  ? 

Ans.  49°  47'  from  horizontal;  31.12  lbs.;  34.05  lbs. 

10.  A  bar  is  supported  at  a  given  angle  with  the  horizontal  by 
strings  attached  at  the  ends.  Show  geometrically  the  relations  that 
must  be  satisfied  by  the  directions  of  the  strings.  Deduce  also  an 
algebraic  expression  for  the  relation  between  the  angles  made  by 
the  strings  and  the  bar  with  the  horizontal. 

Ans.  Let  a  and  b  =  segments  into  which  the  length  of  the  bar 
is  divided  by  the  center  of  gravity  ;  0,  a,  /3  =  angles  made  with  hor- 
izontal by  the  bar  and  strings.     Then 

a  tan  a  —  b  tan  /3  =  {a  -J-  b)  tan  d. 

11.  A  bar  of  mass  20  lbs.,  whose  center  of  gravity  is  at  its  middle 
point,  is  supported  as  in  the  preceding  example,  its  inclination  to  the 
horizontal  being  30°.  The  cord  attached  at  the  lower  end  being  in- 
clined 6o°  to  the  horizontal,  determine  the  inclination  of  the  other 
cord,  and  the  tensions  in  the  two  cords. 

Ans.  6  =  70°  54';  tensions  =  13.23  lbs.  and  8.66  lbs. 

12.  A  bar  6  ins.  long,  whose  mass  is  2  lbs.  and  whose  center  of 
gravity  is  4  ins.  from  one  end,  rests  inside  a  smooth  hemispherical 
bowl  of  radius  4  ins.  What  is  its  inclination  to  the  horizontal  when 
in  equilibrium,  and  what  are  the  pressures  upon  its  ends  ? 

no.  Equilibrium  of  Connected  Bodies. —  If  two  rigid  bodies  in 
equilibrium  are  connected  in  any  way,  as  by  a  smooth  hinge  or  by 
simple  contact,  the  conditions  of  equilibrium  may  be  applied  to  each 
separately.  The  forces  which  the  bodies  exert  upon  each  other  will 
enter  the  equations  as  unknown  quantities  ;  and  are  related  to  each 
other  in  accordance  with  Newton's  third  law  (Art.  35).  That  is,  A 
and  B  being  any  two  bodies,  the  force  which  A  exerts  upon  B  is 


78  THEORETICAL    MECHANICS. 

equal,  opposite  and  collinear  with  the  force  which  B  exerts  upon  A. 
This  will  be  illustrated  by  the  following  problems. 

I.    Two  heavy  bars  AC  and  BC  (Fig.   58),    connected   by  a 
smooth  hinge  at  C,  are  attached  to  a  fixed  body  by  smooth  hinges 

at  A  and  B.     Determine  the  pres- 

\qXc- A®/        sures  exerted  upon  the  bars  by  the 

^X^N.  yfyB         hinges  A  and  B,  and  the  pressure  of 

^xNv>^^  each  upon  the  other  at  C 

^Sr  Solution. —  In  the  triangle  A B C, 

pIG  _8  denote  the  three  angles  by  A,  B,  C, 

and  the  corresponding  opposite  sides 
by  a,  b>  c;  and  let  c  be  horizontal.     Let  the  center  of  gravity  of  each 
bar  be  at  its  middle  point. 
Let  W  =  weight  of  AC; 
W"  =  weight  of  £<7; 
P'  —  resultant  pressure  upon  A  C  at  A  ; 
H'  ==»  horizontal  component  of  P' ; 
V  ==  vertical  component  of  P'; 
P"  bs=  resultant  pressure  upon  BC  at  B; 
H"  =  horizontal  component  of  P" ; 
V"  =  vertical  component  of  P"  \ 
P=  pressure  exerted  upon  AC  by  BCat  C; 
( — P=  pressure  exerted  upon  BC  by  AC  at  C;) 
H  =  horizontal  component  of  P; 
V  =  vertical  component  of  P. 
(The  positive  direction  is  toward  the  right  for  horizontal  forces  and 
upward  for  vertical  forces. ) 

Three  independent  equations  of  equilibrium  for  A  C  may  be  ob- 
tained as  follows: 

Resolving  horizontally, 

ff>  -f  tf=o.        .         .         .         .     (1) 

Resolving  vertically, 

V  +   V  —  W  =  o.  .         .         .     (2) 

Taking  moments  about  A, 

Vb  cos  A  4-  Hb  sin  A  —  y2  W'b  cos  A  =  o.  .     (3) 

Three  independent  equations  of  equilibrium  for  BC  may  be  ob- 
tained in  a  similar  manner  : 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


79 


H"  —  H  =  o (4) 

V"        V-   W"  =  0.  .         .        .     (5) 

Va  cos  B  —  Ha  sin  2>>  -f  %  W"a  cos  £  =  o.         .     (6) 

These  six  equations  suffice  to  determine  the  six  quantities  H,  V> 
H\   V,  H",  V" .     The  solution  may  proceed  as  follows: 

From  (3)  and  (6),  H  and  V  can  be  found ;  the  remaining  un- 
known quantities  can  then  be  determined  at  once  from  the  other 
four  equations.      The  results  are  as  follows: 


H  = 


V  = 


H'  = 


IV'  +   W" 


2  (tan  A  -h  tan  B) 

W  tan  B  —  W"  tan  A  m 

2  (tan  A  +  tan  B) 
W  +  W" 

2  (tan  A  -f-  tan  B)\ 

(  W  +  W"~)  tan  A  , 


r  =  — 

2  2  (tan  ^ 


tan  i?) 


H"  = 


V" 


w 


w 


2  (tan  A  -f  tan  i>' ) 
•IT*        (  ^;  +   W")  tan  ^ 
2  2  (tan  A  +  tan  i? ) 


II.  A  heavy  bar,  AB  (Fig.  59),  is  supported  at  A  by  a  smooth 
hinge  and  leans  against  a  smooth  circular  cylinder  which  rests  in  the 
angle  between  a  horizontal 
floor  and  a  vertical  wall. 
Determine  completely  all 
forces  acting  upon  the  bar 
and  upon  the  cylinder. 

Solution. — Let  the  known 
data  be  the  weight  of  the 
bar,  W  lbs. ;  the  weight  of 
the  cylinder,  W  lbs.;  the 
distance  of  the  center  of 
gravity  of  the  bar  from  A,  a 
ft. ;  the  radius  of  the  cylin- 


8o  THEORETICAL    MECHANICS. 

der,  r  ft. ;  the  distance  AC,  b  ft.     The  angle  BA C  is  fixed  by  the 
known  data  ;  in  fact,  if  BA  C  =  6, 

tan  (0/2)  =  r/(b  —  r). 

The  unknown  forces  are  the  pressure  on  the  bar  at  A,  unknown 
in  magnitude  and  direction  ;  the  pressure  upon  the  bar  at  its  point  of 
contact  with  the  cylinder  (and  the  equal  and  opposite  pressure  upon 
the  cylinder),  known  in  direction  but  not  in  magnitude ;  and  the 
pressures  upon  the  cylinder  at  its  points  of  contact  with  the  floor 
and  wall,  known  in  direction  but  not  in  magnitude. 

Let  Q  denote  the  magnitude  of  the  force  exerted  by  the  cylinder 
upon  the  bar  at  E ;  P  the  magnitude  of  the  force  exerted  upon  the 
bar  by  the  hinge  at  A  ;  H  and  V  the  horizontal  and  vertical  compo- 
nents of  P.  Three  independent  equations  of  equilibrium  for  the  bar 
may  be  written  as  follows  : 

Taking  moments  about  A,  remembering  that  AE  =  AF = 
b  —  r,  we  have 

Q(b  —  r)  —  Wa  cos  0  =  o.     .         .         .     (i) 

Resolving  horizontally, 

H  —  Q  sin  B  =  a     .  .  .  .     (2) 

Resolving  vertically, 

V  +  Q  cos  6  —  W  =  o.        .  .  .      (3) 

These  three  equations  suffice  for  the  determination  of  Q,  H  and  V 


Thus,  from  (1), 

Q 

_  Wa  cos  6 
b  —  r 

From  (2), 

H  = 

-Q 

sin  6  = 

Wa 

sin 
b~ 

0COS  0 

-  r 

From  (3), 

V-- 

=  W 

— 

gcos  e 

=  w\i 

a  cos 
*  — 

>01 

r  _ 

Turning  now  to  the  cylinder,  let  i£  denote  the  horizontal  pres- 
sure exerted  by  the  wall  and  S  the  vertical  pressure  exerted  by  the 
floor.  Since  the  lines  of  action  of  the  four  forces  acting  upon  the  cyl- 
inder intersect  in  a  point,  only  two  independent  equations  can  be 
written.     These  may  be  the  following  :    Resolving  horizontally, 

Q  sin  0  —  R  =  o (4) 

Resolving  vertically, 

—  Q  cos  6  -f  5  —  W  =  o.     c         .         .     (5) 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


81 


From  (4), 
From  (5), 


R  =  Q  sin  0  = 


Wa  sin  6  cos  0 


S  =  W  +  Q  cos  0  =  W  + 


b  —  r 

Wa  cos2  6 


Geometrical  solution. — The  lines  of  action  of  the  three  forces 
acting  upon  the  bar  (P,  Q  and  W)  must  meet  in  a 
point  ;  this  point  is  found  by  prolonging  the  known 
lines  of  action  of  W  and  Q.  Since  W  is  known,  the 
force-triangle  for  the  three  forces  can  be  drawn,  and 
is  shown  in  Fig.  60.  The  forces  P  and  Q  are'  thus 
determined. 

The  forces  acting  upon  the  cylinder  are  four  in 
number  :  a  force  of  magnitude  Q>  opposite  to  the  force 
Q  acting  upon  the  bar;  a  force  W  completely  known; 
forces  R  and  5,  whose  lines  of  action  are  known.  The 
four  being  in  equilibrium,  their  force-polygon  must 
close,  and  can  be  completely  drawn  from  the  data  now 
known  t  is  the  quadrilateral  whose  sides  are  marked 
W\  0i  -  and  R  in  Fig.  60. 

In  any  numerical  case,  the  figure  may  be  drawn  to 
scale  and  the  magnitudes  of  the  required  forces  found 
by  measurement  from  the  figure.  Or,  the  angles  may 
all  be  determined  from  the  given  data,  and  the  lengths 

of  the  sides  representing  the  magnitudes 
of  the  required  forces  may  be  computed 
trigonometrically. 
! 

Examples. 


ra- 


Fig.  60. 


1.  In  Fig.  59,  let  A  C  ='20  ins., 
dius  of  cylinder  =  8  ins.,  AD  =  7  ins., 
W=  12  lbs.,  W  =  20  lbs.  Deter- 
mine all  the  forces  acting  upon  the  bar 
and  cylinder  (a)  algebraically  and  (£) 
geometrically. 

2.  Find  completely  the  forces  acting 
upon  the  bar  and  cylinder  in  Fig.  61, 
the  bar  being  hinged  at  A,  and  the  sur- 
faces of  contact  all  being  smooth.     Let 


82  THEORETICAL    MECHANICS. 

the  weight  of  the  bar  be  10  lbs. ;  the  weight  of  the  cylinder  15  lbs. ; 
the  radius  of  the  cylinder  12  ins.;  CA  (vertical)  24  ins.;  distance  of 
center  of  gravity  of  bar  from  A,  8  ins. 

Ans.  Pressure  between  bar  and  cylinder  =  1.924  lbs. 

3.  Three  uniform  bars  AB,  BC,  CD,  of  equal  length  and  mass, 
are  connected  by  smooth  hinges  at  B  and  C  AB  is  hinged  to  a 
fixed  support  at  A,  and  CD  to  a  fixed  support  at  D.  AB  and  CD 
are  horizontal  and  BC  is  vertical,  the  system  being  held  in  equilib- 
rium by  a  vertical  string  attached  to  A B  at  a  point  two-thirds  the 
distance  from  A  to  B.      Determine  all  forces  acting  upon  each  bar. 

4.  Two  uniform  bars,  equal  in  length  and  mass,  connected  by  a 
smooth  hinge,  rest  upon  a  smooth  cylinder  whose  axis  is  horizontal. 
Determine  the  position  of  equilibrium,  and  the  magnitude  and  direc- 
tion of  every  force  acting  upon  each  bar. 

in.  Simple  Machines. —  It  is  often  desired  to  exert  upon  some 
body  a  force  of  such  magnitude  or  direction  that  means  are  not 
available  for  applying  it  directly.  In  such  a  case  it  may  be  possible 
to  accomplish  the  desired  object  by  indirect  means.  Any  body  or 
system  of  connected  bodies  by  which  such  an  object  is  accomplished 
is  called  a  machine* 

The  bodies  constituting  a  machine  are  usually  either  actually 
or  nearly  in  equilibrium.  The  relations  subsisting  among  the  ap- 
plied forces  may  therefore  be  determined  from  the  principles  of 
Statics. 

A  machine  is  made  to  accomplish  its  object  by  applying  to  it 
forces  tending  to  give  it  a  certain  definite  motion.  A  force  which  aids 
this  motion  is  called  an  effort,  and  a  force  which  opposes  it  a  resist- 
ance. A  force  which  merely  guides  the  motion  without  any  tendency 
to  accelerate  or  to  retard  it,  is  a  constraint. 

The  equal  and  opposite  reaction  to  the  force  which  the  machine 
is  designed  to  exert  is  a  resistance  ;  and  since  the  overcoming  of 
this  resistance  is  the  primary  object  of  the  machine,  it  is  called  a 
useful  resistance.  The  useful  resistance  may  be  called  the  load.  To 
determine  the  relation  between  the  effort  and  the  load  is  the  funda- 
mental statical  problem  presented  by  a  machine. 

Besides  the  useful  resistance  there  are  other  forces  which  act  in 
opposition  to  the  effort  and  diminish  the  effectiveness  of  the  machine 

*  Machines  considered  as  devices  for  the  transmission  of  energy  are 
treated  in  Chapters  XVII  and' XXIII. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


83 


in  overcoming  the  useful  resistance.  These  may  be  called  wasteful 
resistances.  The  most  important  wasteful  resistances  are  the  fric- 
tional  forces  exerted  by  the  bodies  by  whose  agency  the  motion  of 
the  machine  is  guided  in  the  desired  manner.  In  the  present  dis- 
cussion friction  will  be  disregarded.  The  method  of  estimating  its 
effect  will  be  considered  in  Chapter  VII. 

A  machine  is  said  to  yield  a  mechanical  advantage  if  the  load  is 
greater  than  the  effort.  If  the  load  is  less  than  the  effort  there  is  a 
mechanical  disadvantage. 

The  method  of  applying  the  principles  of  equilibrium  to  machines 
is  illustrated  by  the  following  simple  examples.  Other  examples  are 
given  in  later  chapters. 

Lever. — A  lever  is  a  rigid  bar,  movable  about  a  fixed  axis.  The 
effort  and  load  may  be  applied  at  any  points  of  the  bar  and  in  any 
directions,  but  are  usually  taken  as  acting  in  a  plane  perpendicular  to 
the  fixed  axis. 

The  fixed  axis  or  support  about  which  the  lever  turns  is  called 
the  fulcrum.  The  portions  of  the  bar  between  the  fulcrum  and  the 
points  of  application  of  the  effort  and  the  load  are  called  the  arms. 

The  lever  may  be  either  straight  or  bent;  and  the  forces  applied 
to  it  may  or  may  not  be  parallel. 

The  application  of  the  principles  of  equilibrium  to  the  lever  is  so 
simple,  when  friction  is  neglected, 
that  the  solution  of  the  problem  ^ 

will  be  omitted. 

Fixed  Pulley. —  A  pulley  con- 
sists of  a  wheel  which  can  rotate 
freely  about  an  axis,  together  with 
a  flexible  cord  wrapped  about  some 
part  of  the  circumference  of  the 
wheel.  The  axis  about  which  the 
wheel  revolves  may  or  may  not  be 
stationary;  so  that  pulleys  may  be 
classed  as  fixed  and  movable. 

Fig.  62  represents  a  fixed  pul- 
ley, the  effort  and  load  being  ap- 
plied to  the  same  cord,  passing  around  a  portion  of  the  circumfer- 
ence of  the  wheel.     The  relation  between  the  effort  (P)  and  the  load 
(  W)  may  be  found  by  applying  the  principles  of  equilibrium  to  the 


Fig.  62. 


84 


THEORETICAL    MECHANICS. 


system  of  bodies  *  consisting  of  the  wheel  and  the  adjacent  portion 
of  the  cord.  At  A  there  is  a  tension  in  the  cord  equal  to  P,  and 
at  B  a  tension  equal  to  W.  The  only  other  force  acting  upon  the 
system  considered  is  that  exerted  by  the  axle  about  which  the  wheel 
revolves;  and  if  there  is  no  friction  between  the  wheel  and  the  axle, 
this  force  must  act  in  aline  through  the  center  of  the  wheel.  Taking 
moments  about  this  center,  we  have  (calling  a  the  radius  of  the 
wheel ) 

Pa  =  Wa ;  or  P  =  W. 

Let  R  be  the  pressure  exerted  by  the  axle  upon  the  wheel;  then 
R  must  be  equal  and  opposite  to  the  resultant  of  P  and  IV.  Its 
line  of  action  therefore  bisects  the  angle  between  the  two  straight 
portions  of  the  cord;  or,  calling  this  angle  0,  we  have 

R  =  2  W  cos  (0/2). 

Evidently  there  is  no  mechanical  advantage  in  this  case;  the  fixed 
pulley  merely  serves  as  a  means  of  changing  the  direction  of  appli- 
cation of  a  force. 

Movable  Pulley. — A  simple  movable  pulley  is  represented  in 
Fig.  63.  The  relation  between  the  effort  (P)  and  the  load  ( IV)  is 
determined  by  considering  the  system  :on- 
sisting  of  the  wheel  and  the  adjacent  por- 
tions of  the  two  cords,  limited  by  the  points 
A,  B  and  C.  Neglecting  friction  and  the 
weight  of  the  pulley,  the  principles  of  equi- 
librium lead  to  the  following  results : 

Taking  moments  about  the  center  of 
the  pulley,  it  is  seen  that  the  tensions  in 
the  cord  at  A  and  B  are  equal  and  each 
equal  to  P. 

Resolving  forces  vertically,  it  is  seen 
that  the  tension  in  the  cord  at  C  (which  is 
equal  to  W)  must  be  equal  to  the  sum  of 
the  tensions  at  A  and  B,  or  2P;  that  is, 

P=  WI2. 


*  Although  we  have  thus  far  applied  the  principles  of  equilibrium  only  to 
rigid  bodies,  it  is  allowable  to  treat  as  rigid  any  body  or  connected  system  of 
bodies,  every  part  of  which  is  in  equilibrium.     See  Chapter  VI. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


85 


Sy 'Stems  of  Pulleys. — By  combining  fixed  and  movable  pulleys  in 
various  ways,  the  mechanical  advantage  may  be  made  very  great, 
being  limited  only  by  the  prejudicial  resistances  due  to  friction  and 
the  lack  of  perfect  flexibility  of  the  cords. 

Fig.  64  shows  a  system  very  commonly  employed  in  rais- 
ing heavy  weights.  Two  sets  of  wheels  are  employed  (shown  at 
A  and  £),  each  set  consisting  of  several  pulleys  mounted  on  a 
common  axis  but  revolving  independently  of  one  another.  One 
set  (A )  is  attached  to  a  fixed  support,  while  to  the  other  set, 
which    is    movable,    the    load    is    applied.* 

Neglecting  friction  and  the  rigidity  of  the 
cord,  the  relation  between  P  and  W  is  easily 
obtained.  The  tension  has  the  same  value  in 
all  parts  of  the  cord,  as  may  be  seen  by  apply- 
ing the  principle  of  moments  to  each  wheel 
separately,  taking  origin  at  the  center.  If  n 
is  the  number  of  "plies,"  or  portions  of  rope 
supporting  the  lower  set  of  pulleys,  we  have 
(neglecting  the  weight  of  the  pulleys  and  of 
the  cord,  and  assuming  that  the  straight  por- 
tions of  the  cord  are  all  parallel) 

nP=W-     P=W\n. 


Examples. 

1.  The  Dar  AB  (Fig.  65)  is  free  to  rotate 
about  a  smooth  pin  at  C.     The  effort  P  and    R 
the  resistance  W  act  at  angles  d  and  </>  with   ** 


C 


e 


Fig.  65. 


*  In  the  system  of  pulleys  as  actually  constructed,  the  wheels  of  each  set 
are  usually  of  equal  diameter  and  mounted  side  by  side  upon  the  same  axle. 
The  different  portions  of  the  cord  are  therefore  not  all  coplanar.  Each  of 
the  separate  bodies  (each  pulley  and  the  block  carrying  the  pulleys)  is  how- 
ever acted  upon  by  forces  which  are  practically  coplanar. 


86 


THEORETICAL   MECHANICS. 


the  bar.    Determine  the  relation  between  P  and  W,  and  compute  the 
magnitude  and  direction  of  the  force  exerted  upon  the  bar  by  the  pin. 

2.  In  Ex.  i,  let  AC  =3  ft-,  CB  =  2  ft.,  6  =  6o°,  $  =  3o°, 
W  =  85  lbs.     Determine  P  and  the  magnitude  and  direction  of  the 

pressure  exerted  upon  the  pin. 

Ans.  P  —  32.7  lbs. 

3.  If  P  and  W  (Fig.  62)  act  in  direc- 
tions inclined  90°  to  each  other,  what  is 
the  pressure  on  the  axle  ? 

4.  In  case  of  a  movable  pulley,  let 
the  effort  act  at  an  angle  a  with  the  hori- 
zontal. What  must  be  the  direction  of 
the  cord  between  the  pulley  and  the 
fixed  point  of  attachment  of  the  cord  ? 
What  is  the  relation  between  Pand  W ? 

Ans.  P=  W/2sma. 

5.  Determine  the  relation  between  P 
and  W  (  Fig.  63)  taking  account  of  the 
weight  of  the  pulley. 

6.  Determine  the  relation  between  P 
and  W  in  case  of  a  system  such  as 
shown  in  Fig.  64,  the  number  of  mov- 
able pulleys  being  3. 

7.  In  Fig.  64,  what  must  be  the 
strength  of  the  cord  if  a  weight  of  5,000 
lbs.  is  to  be  raised  ? 

8.  In  Fig.  66,  let  the  weight  W  be 
1,000  lbs.      Determine  the  tension  in  each  cord  and  the  value  of  P. 

9.  Solve  Ex.  8,  taking  into  account  the  weights  of  the  pulleys, 
5  lbs.  each. 

112.  Balance. —  Since  the  weights  of  bodies  in  the  same  locality 
are  proportional  to  their  masses,  the  mass  of  any  body  may  be  deter- 
mined by  comparing  its  weight  with  that  of  a  body  whose  mass  is 
known. 

The  essential  parts  of  a  balance  for  comparing  the  weights  of 
bodies  are  shown  in  Fig.  67.  The  beam  L  can  turn  freely  in  a  ver- 
tical plane  about  a  fixed  point  C.  At  A  and  B  are  suspended  the 
"  scale-pans  ' '  M  and  N,  upon  which  are  placed  the  bodies  whose 
weights  are  to  be  compared.  The  center  of  gravity  of  the  whole 
apparatus  *  is  at  some  point  D.     When  the  scale-pans  are  vacant, 

*  In  determining  this  center  oi  gravity,  the  scale-pans  are  to  be  regarded 
as  particles  located  at  their  points  of  suspension  A  and  B. 


Fig.  66. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES. 


87 


the  point  D  will  be  vertically  below  C  in  the  position  of  equilibrium. 
In  this  position  let  CA  and  CB  make  equal  angles  with  the  horizon- 
tal, and  let  the  lengths  A  C  and  CB  be  each  equal  to  a.  If  bodies  of 
equal  mass  (and  weight)  are  placed  in  the  two  scale-pans,  the  position 
of  equilibrium  will  be  unchanged. 

Let  a  body  of  mass  P  be  placed  at  M,  and  a  body  of  mass  Q  at 
N,  P  being  greater  than  Q,  and  let  the  beam  assume  a  position  of 
equilibrium,  6  being  the  angle  between  CD  and  the  vertical  in  this 
position.  If  initially  the  lines  CA  and  CB  make  with  the  horizontal 
the  angle  a,  their  inclinations  to  the  horizontal  in  the  new  position 


Fig.  67. 


of  equilibrium  are  a  +  0  and  a  —  6  respectively.    Let  CD  =  h,  and 
let  G  =  weight  of  balance.     Taking  moments  about  C, 


Pa  cos  (a 
Solving  for  0, 


0) 


tan  6 


Qa  cos  (a  —  6)  —  Gh  sin  6  =  o. 
(P —  Q)a  cos  a 


(P+  Q)a  sin  a  +  Gh 


In  order  that  a  small  difference  between  P  and  Q  may  be  detected 
with  certainty,  a  small  value  of  P  —  Q  should  cause  a  relatively  great 
value  of  6.  This  sensitiveness  depends  upon  the  values  of  a,  h  and  a. 
For  given  values  of  a  and  a,  the  sensitiveness  increases  as  h  de- 
creases. For  given  values  of  a  and  h,  the  sensitiveness  increases 
as  a  decreases. 


88  THEORETICAL    MECHANICS. 

A  very  high  degree  of  sensitiveness  can  be  obtained  only  at  the 
expense  of  stability.  That  is,  if  a  very  small  value  of  P  —  Q  pro- 
duces a  large  value  of  0,  the  beam  cannot  readily  be  brought  into  a 
condition  of  equilibrium,  but  will  oscillate  between  positions  far  from 
the  position  of  rest. 

In  the  practical  use  of  the  balance,  the  body  whose  mass  is  to  be 
determined  is  placed  in  one  scale- pan,  and  standard  "weights" 
(bodies  of  known  mass)  are  placed  in  the  other  in  sufficient  quantity 
so  that  the  position  of  equilibrium  is  the  same  as  when  both  scale- 
pans  are  vacant.  In  order  that  the  indications  of  the  balance  may 
be  correct  when  used  in  this  manner,  the  horizontal  projections  of 
AC  and  BC  (Fig.  67)  must  be  equal.  These  horizontal  projections 
may  be  called  the  arms  of  the  balance. 

If  the  arms  are  unequal,  this  fact  may  be  detected  by  interchanging 
the  weights  in  the  pans.  Thus,  let  a  body  whose  true  weight  is 
x  lbs.  be  balanced  against  standard  weights  in  a  balance  whose  arms 
are  a  and  b.  Let  the  apparent  weight  of  x  be  P  lbs.  when  it  acts 
with  the  arm  a,  and  Q  lbs.  when  it  acts  with  the  arm  b.     Then 

xa  —  Pb\   xb  =  Qa. 
From  these  equations, 


x=yPQ;      a/b=Vp/Q. 


Examples. 

1.  In  Fig.  67,  let  the  distances  AC  and  CB  be  each  15  ins., 
CD  6  ins.,  the  angles  ACD  and  BCD  each  8o°,  and  the  weight 
of  the  balance  4  lbs.  What  is  the  position  of  equilibrium  if  weights 
of  10  lbs.  and  10  lbs.  -J-  }i  oz.  are  placed  in  the  two  scale-pans? 

Ans.  6  =  0°  5'+. 
2.  A  body  appears  to  weigh  18  oz.  when  placed  in  one  scale-pan 
and  18^  oz.  when  placed  in  the  other.     What  is  its  true  weight,  and 
what  is  the  ratio  between  the  lengths  of  the  arms  ? 

Ans.  18.248  oz.;  0.986. 

Miscellaneous  Examples. 

1.  A  bar  AB  rests  with  the  end  A  upon  a  smooth  horizontal 
plane,  and  leans  against  a  smooth  cylindrical  peg  at  a  given  distance 
above  the  plane.  It  is  held  from  slipping  by  a  horizontal  string  A  C 
attached  at  a  fixed  point  C.  Determine  all  forces  acting  upon  the 
bar. 


EQUILIBRIUM    OF    NON-CONCURRENT    FORCES.  89 

2.  In  Ex.  i,  let  the  height  of  the  peg  above  the  plane  be  18 
ins.,  the  distance  of  the  center  of  gravity  of  the  bar  from  A  12 
ins.,  the  inclination  of  the  bar  to  the  plane  25°,  its  weight  35  lbs. 
Determine  all  forces  acting  upon  the  bar. 

Ans.  Vertical  pressure  at  A  =  26.90  lbs.;  pressure  of  peg  = 
8.94  lbs.;  tension  =  3.78  lbs. 

3.  Let  the  data  be  as  in  Ex.  1,  except  that  the  horizontal  string 
A  C  passes  over  a  smooth  pulley  at  C  and  sustains  a  body  of  known 
weight.  Determine  the  position  of  equilibrium  and  all  forces  acting 
upon  the  bar.      [The  solution  leads  to  a  cubic  equation.] 

4.  Solve  Ex.  3  with  the  following  numerical  data :  Weight  of  bar, 
10  kilogr. ;  weight  of  suspended  body,  2  kilogr. ;  height  of  peg  above 
plane,  50  cm. ;  distance  of  center  of  gravity  of  bar  from  A,  50  cm. 

Ans.   Angle  of  bar  with  horizontal  =  770  56'  or  28°  30'. 

5.  A  uniform  bar  AB  of  known  mass  and  length  is  supported  in 
a  horizontal  position  by  a  smooth  hinge  at  A  and  a  cord  BC  inclined 
to  the  bar  at  a  known  angle  and  fixed  at  C.  Two  bodies  of  known 
mass  are  suspended  from  the  bar  at  points  midway  between  the 
center  of  gravity  and  the  ends  A  and  B  respectively.  Determine  all 
forces  acting  on  the  bar. 

6.  Solve  Ex.  5  with  the  following  numerical  data  :  Weight  of  bar, 
18  lbs.;  suspended  weights,  3  kilogr.  and  5  kilogr.;  inclination  of 
string  to  bar,  37°  (measured  from  prolongation  of  AB). 

Ans.  Tension  =  14.26  kilogr.;  hinge  reaction  =  13.67  kilogr. 
inclined  1460  20'  to  AB. 

7.  A  bar  AB  of  weight  W  is  supported  by  a  smooth  hinge  at  A 
and  a  string  attached  at  B.  The  string  passes  over  a  smooth  peg  at 
C and  supports  a  body  of  weight  P.  ACis  horizontal  and  equal  to 
AB,  and  the  center  of  gravity  of  the  bar  is  at  its  middle  point.  De- 
termine the  position  of  equilibrium,  and  all  forces  acting  on  the  bar. 

CAB        P 

Ans.  cos  — —  = 

2  2W 


*.V 


2W'1 


Pl 


8.  Solve  Ex.  7,  taking  W  =  28  lbs.,  P=  6  kilogr. 

Ans.  CAB  =  210  54  or  2410  20'.  Hinge  reaction  =  15.25  lbs. 
or  36.53  lbs. 

9.  In  Ex.  7,  let  AC  and  AB  be  unequal,  A  C  being  still  horizontal. 
Deduce  a  cubic  equation  for  determining  cos  0,  where  6  is  the  angle 
between  the  bar  and  the  horizontal. 

10.  Solve  Ex.  9,  taking  AC=  2  {AB),  W=  P  =  12  kilogr. 

Ans.  6  =  14°  55',  or  2 1 50  42'. 

11.  Modify  Ex.   7  by  taking  AC  =  AB,  but  not  horizontal. 

12.  A  smooth  cylinder  of  radius  a  is  placed  with  its  axis  hori- 
zontal and  parallel  to  a  smooth  vertical  wall  at  distance  h.  A  uni- 
form bar  of  weight  W  and  length  2/  rests  against  the  cylinder  and 


90  THEORETICAL    MECHANICS. 

with  its  lower  end  against  the  wall.      Determine  the  position  of  equi- 
librium. 

Ans.  If  0  =  angle  between  bar  and  vertical,  /  sin3  0  -f-  a  cos  0 
=  k. 

13.  In  Ex.  12,  let  the  radius  of  the  cylinder  be  very  small 
compared  with  h  and  /.  Solve  with  the  following  data :  W  =  15 
kilogr. ,  h  =  6  ins. ,  /  =  12  ins.  Determine  the  position  of  equilibrium 
and  all  forces  acting  on  the  bar. 

14.  A  uniform  smooth  bar  of  length  2/  and  weight  W  rests  in  a 
hemispherical  bowl  of  radius  a.  Assuming  the  length  to  be  so  great 
that  the  upper  end  projects  beyond  the  edge  of  the  bowl,  determine 
the  position  of  equilibrium  and  all  forces  acting  on  the  bar. 

Ans.  The  angle  0  between  the  bar  and  the  horizontal  is  given  by 
the  equation  cos  0  =  (/  ±  V P  +  320 2)/ 8a. 

15.  Solve  Ex.  14  with  the  following  numerical  data:  W  =  12 
kilogr. ,  /  =  o.  45  met. ,  a  =  o.  40  met. 

Ans.  0  =  30°  28';  pressure  at  lower  end  =  7.05  kilogr. 

16.  A  heavy  bar  is  supported  by  a  string  attached  at  any  two 
points  and  passing  over  a  smooth  peg.  Determine  the  position  of 
equilibrium  and  the  tension  in  the  string. 

Ans.  The  peg  divides  the  string  into  segments  whose  lengths  are 
directly  proportional  to  the  distances  of  the  center  of  gravity  of  the 
bar  from  the  points  of  attachment  of  the  string. 

17.  In  the  preceding  example  let  the  weight  of  the  bar  be  12 
lbs.,  the  distances  of  the  center  of  gravity  from  the  points  of  attach- 
ment of  the  strings  2  ft.  and  3  ft.,  the  length  of  the  string  7.5  ft. 
Determine  the  tension  in  the  string  in  the  position  of  equilibrium. 

Ans.  7.89  lbs. 

18.  What  horizontal  force  is  necessary  to  pull  a  carriage  wheel 
over  a  smooth  obstacle,  the  radius  of  the  wheel  being  a,  its  weight 
W,  and  the  height  of  the  obstacle  h  ? 

19.  A  heavy  uniform  beam,  movable  in  a  vertical  plane  about  a 
smooth  hinge  at  one  end,  is  sustained  by  a  cord  attached  to  the  other 
end.  The  angle  between  the  bar  and  the  vertical  being  fixed,  deter- 
mine what  direction  of  the  cord  will  cause  the  least  pressure  on  the 
hinge.  Determine  the  corresponding  values  of  the  unknown  forces 
acting  on  the  bar. 

Ans.  Let  a  =  angle  of  bar  with  vertical,  0  =  angle  of  string  with 
bar,  then  tan  0  =  }4  tan  a. 

20.  A  heavy  beam  rests  with  one  end  against  a  smooth  inclined 
plane  ;  to  the  other  end  is  attached  a  cord  which  passes  over  a  smooth 
pulley  and  sustains  a  given  weight.  Determine  the  position  of  equi- 
librium. 

21.  The  roof-truss  shown  in  skeleton  in  Fig.  68  is  supported  by 
a  smooth  horizontal  surface  at  A  and  a  smooth  hinge  at  B.      It  is  re- 


EQUILIBRIUM    OF    NON-CONCURRENT   FORCES. 


91 


quired  to  sustain  the  wind  pressure  on  a  portion  of  the  roof  of  width 
1 2  ft.  and  length  A  C  or  CB.     Assume  the  pressure  to  be  normal  to 
the  roof  and  uni- 
formly   equal    to 
20  lbs.  per  sq.  ft. 
Compute  the  sup- 
porting forces  at 
A   and   B   when 
the  wind  is  from  j\_ 
the    right  ;    also 
when   it  is   from 
the  left. 

22.  Take  data 

as  in  Ex.  17,  but  assume  the  peg  rough,  so  that  the  tensions  in  the 
two  portions  of  the  string  may  be  unequal.  If  the  bar  is  in  equi- 
librium when  the  peg  divides  the  string  into  segments  1.9  ft.  and  5.6 
ft.  in  length,  what  are  the  tensions  in  the  two  portions  of  the  cord  ? 

Ans.   10.46  lbs.  and  4. 18  lbs. 

23.  The  lines  of  action  of  three  forces  intersect  in  points  A,  By 
and  C.  Prove  that,  if  their  resultant  is  a  couple,  they  are  propor- 
tional to  the  vectors  AB,  BC,  CA. 

24.  A  uniform  bar  of  weight  W  rests  with  its  ends  against  two 
smooth  planes  whose  inclinations  to  the  horizontal  are  a  and  /3,  that 
of  the  bar  being  6.  A  weight  P  is  suspended  from  a  point  distant 
one-third  the  length  of  the  bar  from  one  end.  Required  the  value 
of  P  for  equilibrium. 


CHAPTER  VI. 

EQUILIBRIUM    OF    PARTS    OF    BODIES    AND    OF   SYSTEMS    OF    BODIES. 

§  i.  Equilibrium  of  Any  Part  of  a  Body. 

113.  External  and  Internal  Forces  Acting  on  a  Body. —  The 

conditions  of  equilibrium  have  thus  far  been  applied  only  to  a  par- 
ticle or  to  a  rigid  body  regarded  as  a  whole.  In  case  two  or  more 
bodies  were  considered,  each  was  regarded  as  presenting  a  separate 
problem  in  equilibrium.  For  certain  purposes,  however,  it  is  desir- 
able to  confine  the  attention  to  a  portion  of  a  rigid  body;  and  for 
other  purposes  it  is  found  convenient  to  treat  two  or  more  bodies 
as  together  forming  a  system.  As  a  preliminary  to  the  discussion  of 
problems  from  these  points  of  view,  it  is  useful  to  recur  to  the  funda- 
mental conception  of  force. 

By  the  definition  (Art.  32),  a  force  is  an  action  exerted  by  one 
body  upon  another.  Again,  by  a  fundamental  law  (Art.  35),  forces 
always  act  in  pairs  ;  so  that  when  any  one  body  exerts  a  force  upon 
another,  the  second  body  exerts  an  equal  and  opposite  force  upon 
the  first.  Thus  every  force  concerns  two  bodies  ;  although  in  the  pre- 
ceding chapters  the  attention  has  in  every  case  been  directed  espe- 
cially to  a  single  body, — that  upon  which  the  force  acts,  —  and  the 
body  exerting  the  force  has  not  always  been  specified. 

The  word  body,  in  the  definition  of  force,  must  be  understood  to 
mean  any  portion  of  matter.  The  two  portions  may  be  separate 
from  each  other  (as  in  most  of  the  cases  hitherto  discussed),  or  they 
may  be  parts  of  a  single  body.  This  leads  to  the  following  classifi- 
cation of  the  forces  applied  to  a  body: 

An  external  force  is  one  exerted  upon  the  body  in  question  by 
some  other  body. 

An  internal  force  is  one  exerted  upon  one  portion  of  the  body  by 
another  portion  of  the  same  body. 

114.  Conditions  of  Equilibrium  for  a  Rigid  Body  as  a  Whole. — 
The  discussions  of  the  composition,  resolution  and  equilibrium  of 
forces,  as  thus  far  given,  have  related  to  external  forces  applied  to 
the  same  rigid  body,  internal  forces  not  appearing  in  the  equations 
or  conditions  of  equilibrium. 


EQUILIBRIUM    OF    PARTS    OF    BODIES.  93 

Thus,  although  it  is  generally  true  that  adjacent  portions  of  a 
rigid  body  exert  forces  upon  each  other,  these  forces  do  not  come  into 
consideration  when  the  problem  relates  solely  to  the  equilibrium  of 
the  body  as  a  whole.  It  may,  however,  be  desired  to  gain  some  in- 
formation regarding  these  internal  forces.  It  will  then  be  necessary 
to  direct  the  attention  to  a  portion  of  the  body,  and  to  consider  the 
relations  among  all  the  forces  which  act  upon  this  portion. 

115.  Conditions  of  Equilibrium  for  a  Portion  of  a  Body. — 
Since  the  body  to  which  the  conditions  of  equilibrium  apply  may  be 
any  portion  of  matter  whose  particles  are  rigidly  connected,  it  may 
be  any  connected  portion  of  a  rigid  body.  But  in  writing  the  equa- 
tions of  equilibrium  for  a  portion  of  a  body,  certain  forces  must  be 
included  which  are  omitted  if  the  equilibrium  of  the  whole  body  is 
considered.  These  are  the  forces  exerted  upon  the  portion  in  ques- 
tion by  other  portions  of  the  same  body.  Such  forces  are  internal 
when  the  whole  body  is  considered,  but  external  when  one  part  is 
considered. 

To  illustrate,  let  AB  (Fig.  69)  represent  a  bar  acted  upon  by  two 
forces  of  equal  magnitude  applied  at  the  ends  in  opposite  directions 
parallel  to  the  length  of  the  bar.  These  forces  are  exerted  upon  AB 
by  some  other  bodies  not  specified.  If  the  whole  bar  be  considered, 
the  external  forces  are  the  two  forces  named.  But  suppose  the  body 
under  consideration  is  AC,  a  portion  of  AB  ;  the  external  forces  act- 
ing upon  this  body  are  ( 1 )  a  force  ap- 
plied at  A  (already  mentioned)  and  A 
(2)  a  force  applied  at  C  (exerted  upon 


AC  by  OB).     This  latter  force  is  in-  Fig.  69. 

ternal  to  the  bar  AB  but  external  to 

AC      In  applying  the  conditions  of  equilibrium  to  AB,   the  only 

forces  to  enter  the  equations  are  those  exerted  at  A    and  B  as 

already  mentioned  ;  but  if  the  equations  of  equilibrium  for  A  C  are 

to  be  written,  the  forces  to  be  included  are  the   force  at  A  and  the 

force  exerted  upon  AC  by  CB  at  C 

§  2.  Determination  of  Internal  Forces. 

116.  General  Method  of  Determining  Internal  Forces. —  Little 
is  in  general  known  regarding  the  forces  which  contiguous  portions 
of  a  body  exert  upon  each  other.     But  when  the  external  forces  are 


94 


THEORETICAL   MECHANICS. 


known,  the  internal  forces  can  be  partly  determined  by  applying  the 
conditions  of  equilibrium  to  portions  of  the  body.  Thus,  to  take  the 
simplest  and  most  important  case,  suppose  it  is  known  that  not  only 
the  whole  body,  but  every  portion  of  it,  is  in  equilibrium.*  For 
such  cases  the  following  proposition  may  be  stated: 

If  a  body,  every  part  of  which  is  in  equilibrium,  be  conceived  as 

made  up  of  two  parts,  X  and  Y,  then  the  internal  forces  exerted  by 

X  upon  Y,  together  with  the  external  forces 

acting  on  Y,  form  a  system  in  equilibrium. 

Thus,    by    applying  the  conditions    of 

equilibrium  to  the  body  Y,   the  resultant 

of  all  the  forces  exerted   by  X  upon    Y 

may  be  determined  ;    it  being  equal  and 

opposite  to  the  resultant  of  the  external 

forces  applied  to  Y. 

The  individual  forces  exerted  by  the  particles  of  X  upon  those  of 

Y  cannot,  however,  be  determined,  these  forces  being  in  general  very 

numerous  and  acting  in  various  unknown  directions. 


Fig.  70. 


% 


\l 


Examples. 

1.  A  beam  8  ft.  long,  weighing  40  lbs.  per  linear  foot,  rests  hori- 
zontally upon  supports  at  the  ends.  Dividing  the  beam  by  a  trans- 
verse plane  3  ft.  from  one  end,  determine 
the  resultant  of  the  forces  exerted  by  each 
portion  upon  the  other. 

The  forces  acting  upon  the  left  por- 
tion of  the  beam  (Fig.  71)  are  (1)  the 
supporting  force  of  160  lbs.;  (2)  the 
weight  of  three  linear  feet  of  the  beam, 

equivalent  to  a  force  of  1 20  lbs.  applied  1.5  ft.  from  the  left  end  ; 
and  (3)  the  forces  exerted  by  the  other  portion  of  the  beam.  The 
resultant  of  the  first  two  is  an  upward  force  of  40  lbs.  acting  in  a 
line  4. 5  ft.  to  the  left  of  the  left  support.     Hence  the  forces  exerted  by 


Fig.  71. 


*  It  is  to  be  carefully  noted  that  the  equilibrium  of  a  body  does  not  neces- 
sarily imply  the  equilibrium  of  every  portion  of  it.  Thus,  if  the  external 
forces  applied  to  the  whole  body  are  balanced,  it  may  still  have  a  motion  of 
rotation,  in  which  case  the  forces  acting  upon  a  portion  of  the  body  will  in 
general  be  unbalanced.  This  will  be  apparent  when  the  rotation  of  a  rigid 
body  has  been  studied.  In  very  many  cases,  especially  such  as  arise  in  en- 
gineering practice,  the  bodies  are  not  only  in  equilibrium,  but  at  rest.  In 
such  cases  all  parts  of  the  body  are  also  in  equilibrium. 


EQUILIBRIUM    OF    PARTS    OF    BODIES.  95 

the  other  portion  of  the  beam  have  for  their  resultant  a  downward 
force  of  40  lbs.  acting  in  a  line  4.5  ft.  to  the  left  of  the  left  support. 

2.  A  beam  12  ft.  long,  of  uniform  density  and  cross-section, 
weighing  20  lbs.  per  linear  foot,  rests  horizontally  upon  supports  at 
the  ends.  Conceiving  the  beam  divided  by  a  transverse  plane 
through  the  middle  of  the  length,  determine  the  resultant  of  the 
forces  exerted  by  each  portion  upon  the  other. 

3.  In  Ex.  2,  let  the  dividing  plane  be  3  ft.  from  one  end.  Deter- 
mine the  resultant  of  the  forces  exerted  by  each  portion  upon  the 
other. 

4.  Let  the  same  beam  rest  upon  a  single  support  at  the  middle. 
Determine  the  resultant  of  the  forces  exerted  upon  each  other  by  the 
two  portions  described  in  Ex.  2  ;  also  by  the  two  portions  described 
in  Ex.  3. 

5.  Let  the  same  beam  rest  upon  supports,  one  of  which  is  2  ft. 
from  one  end  and  the  other  4  ft.  from  the  other  end.  Answer  the 
questions  asked  in  examples  2  and  3. 

6.  A  uniform  beam  17  ft.  long,  weighing  120  lbs.,  rests  horizon- 
tally upon  supports  at  the  ends  and  sustains  two  loads  :  a  load  of  40 
lbs.  3  ft.  from  the  left  end,  and  a  load  of  80  lbs.  6  ft.  from  the  right 
end.  Taking  a  transverse  section  4  ft.  from  the  left  end,  compute 
the  resultant  of  the  forces  exerted  by  one  of  the  two  portions  of  the 
beam  upon  the  other. 

Ans.  52. 94  lbs. ;  3.3  ft.  from  left  end  and  20.3  ft.  from  right  end. 

7.  In  Ex.  6,  let  the  transverse  section  be  taken  4  ft.  from  the 
right  end.     Answer  the  same  question. 


Fig.  72. 

117.  Jointed  Frame. —  Engineering  practice  has  to  deal  with 
structures  made  up  of  straight  bars  connected  at  the  ends  by  hinge 
joints.  (See  Art.  42.)  A  structure  thus  made  is  called  a.  jointed 
frame  or  truss.  The  determination  of  the  internal  forces  in  such  a 
frame  is  an  important  problem  in  engineering. 

Fig.  72  represents  in  skeleton  a  simple  form  of  roof  truss,  resting 
upon  supports  at  the  ends.  Such  a  truss  is  acted  upon  by  external 
forces   consisting  of  the  weight  of  its  own  members,  the  weight  of 


96  THEORETICAL    MECHANICS. 

the  roof  proper,  sometimes  the  weight  of  snow  and  the  pressure  of 
wind,  and  the  supporting  forces  at  the  ends.  Each  of  the  bars 
composing  the  truss  is  subjected  to  certain  forces  tending  to  break  it; 
this  tendency  is  resisted  by  the  internal  forces  called  into  action  be- 
tween the  contiguous  portions  of  the  bar.  If  these  internal  forces 
exceed  certain  values,  the  material  is  injured  and  the  bar  may  be 
broken;  it  is  therefore  important  to  be  able  to  determine  their  mag- 
nitudes before  designing  the  members. 

To  solve  this  problem  the  general  method  of  Art.  116  is  em- 
ployed. But  in  order  to  simplify  the  problem  and  make  it  com- 
'pletely  determinate,  certain  assumptions  are  made  which  are  only 
approximately  true. 

(a)  It  is  assumed  that  the  hinges  are  without  friction.  The 
effect  of  this  may  be  seen  by  referring  to  Fig.  14  (p.  21),  which  repre- 
sents a  portion  of  one  bar.  The  connection  with  other  bars  is  made  by 
means  of  a  pin  A  which  passes  through  holes  in  the  ends  of  the  bars. 
The  pin  being  assumed  smooth,  the  pressure  of  the  pin  upon  the  bar 
is  normal  to  the  cylindrical  surfaces  of  both  pin  and  bar,  and  there- 
fore acts  through  the  axes  of  these  cylinders. 

(b)  It  is  assumed  that  all  external  forces  acting  upon  the  truss 
are  applied  to  the  pins.  If  this  is  the  case,  the  only  forces  acting  on 
any  bar  are  those  exerted  by  the  two  pins  at  its  ends.  These  two 
forces  being  in  equilibrium,  must  be  equal  and  opposite  and  must 
have  the  same  line  of  action,  which  is  therefore  the  line  joining  the 
centers  of  the  hinges.  If  a  transverse  section  be  taken  dividing  the 
bar  into  two  parts,  the  principle  of  Art.  1 16  shows  that  the  resultant 
force  exerted  by  each  part  upon  the  other  must  have  the  same  line  of 
action  as  the  forces  exerted  upon  the  ends  of  the  bar  by  the  pins. 

118.  Determination  of   Internal  Forces   in  Jointed  Frame. — 

The  method  of  determining  internal  forces  in  a  jointed  frame  will  now 
be  explained  by  reference  to  a  numerical  problem.  Fig.  73  shows 
the  dimensions  of  a  roof  truss  supporting  three  vertical  loads  at 
upper  joints,  and  supported  upon  horizontal  surfaces  at  the  ends. 
It  is  required  to  determine  the  internal  forces  in  each  of  the  bars. 
No  loads  will  be  considered  except  those  shown.  The  first  step  in 
the  solution  is 

(a)  The  determination  of  the  supporting  forces.  — Assuming  these 
to  be  vertical,  each  must  be  equal  to  half  the  total  load  supported, 
i.e.,  to  375  lbs. 


EQUILIBRIUM    OF    PARTS    OF    BODIES. 


97 


(b)  Determination  of  internal  forces. —  Let  a  section  be  taken, 
as  MNy  dividing  the  truss  into  two  parts,  X  and  Y,  and  let  the 
conditions  of  equilibrium  be  applied  to  either  part,  as  X.     This  body 


X  is  acted  upon  by  an  upward  force  of  375  lbs.  at  the  support ;  a 
downward  load  of  250  lbs.;  and  the  forces  exerted  by  Y.  In  ac- 
cordance with  the  assumptions  made  in  Art.  117,  these  forces  exerted 
by  Y  are  three,  acting  along  the  lines  of  the  three  members  cut  by 
the  section  MN.  The  magnitudes  of  these  forces  are,  however,  un- 
known. If  three  independent  equations  of  equilibrium  be  written 
for  the  body  X,  the  only  unknown  quantities  which  will  enter  them 
are  these  three  unknown  force-magnitudes,  which  may  therefore  be 
determined  by  solving  the  equations. 

Let  Px ,  P2  and  Pz  denote  the  magnitudes  of  the  three  unknown 
forces  acting  upon  X  (Fig.  74),  and  let  them  be  assumed  to  act 


Fig.  74. 


6f 

(X)  D) 

r^* 

A 

/^ 

/ 

/ 

M 
S 
01 

1 

500 


#JE7 


Fig.  75- 


away  from  the  body  X.  (Their  actual  directions  will  then  be  known 
from  the  signs  of  their  numerical  values.)  Let  the  three  equations 
be  written  as  follows  : 

7 


98 


THEORETICAL    MECHANICS. 


Taking  origin  of  moments  at  A,  Px  and  P3  are  eliminated.  The 
arm  of  P2  is  5^/2  ft,  and  the  equation  is 

pi  X  sV2  -250X  10  =  0.     .         .         .      (1) 

Next  taking  origin  of  moments  at  B,  P2  and  P.A  are  eliminated,  and 
the  arm  of  Pl  is  2^/5  ft.     Hence 

—P1  X  2^/5  —  375  X  10  =  o.    .         .         .     (2) 

Again,  taking  moments  about  C,  Px  and  P2  are  eliminated,  and  the 
arm  of  P%  is  10  ft     Hence 

P.,  X  10  +  250  X  10-375X  20  =  0.     .         .     (3) 
Solving  these  equations, 

Px  =  -  838. 5  lbs  ;     P2  =  +  353.6  lbs  ;     P,  =  +  500.0  lbs. 

The  signs  of  these  results  show  that  P2  and  P3  act  away  from  the 
body  Jf,  while  Px  acts  toward  X. 

The  forces  exerted  by  X  upon  Y  are  equal  in  magnitude  to  Px , 
P2  and  P.s ,  but  opposite  in  direction. 

Since  the  independent  equations  of  equilibrium  may  be  written  in 
many  different  ways,  the  details  of  the  solution  may  be  varied  indef- 
initely. It  is  often  advantageous  to  use  resolution  equations  instead 
of  moment  equations.* 

The  following  examples  may  be  solved  by  application  of  the  fore- 
going principles. 

Examples. 

1.   Determine  the  internal  forces  in  all  the  bars  of  the  truss  shown 

in  Fig.  73. 

2.  Let  the  truss 
shown  in  Fig.  76 
sustain  loads  of  400 
lbs.  and  800  lbs. 
due  to  wind  pres- 
sure, acting  nor- 
mally to  the  roof 
and  applied  at  the 
joints  as  shown. 
Suppose    the  truss 


*  In  the  above  discussion  the  aim  has  been  merely  to  exhibit  clearly  the 
genera.'  principle  s  upon  which  the  determination  of  internal  forces  in  framed 
structures  is  based.  For  a  full  treatment  the  reader  must  consult  treatises 
devoted  especially  to  the  subject. 


EQUILIBRIUM    OF    A   SYSTEM    OF    BODIES.  99 

supported  at  one  end  by  a  smooth  horizontal  surface  and  at  the 
other  end  by  a  hinge.  Compute  the  internal  forces  in  all  the  bars  of 
the  truss,  neglecting  all  loads  except  those  specified. 

[First  determine  the  supporting  forces  at  the  ends  by  considering 
the  equilibrium  of  the  truss  as  a  whole.  Then  apply  the  above 
method  to  the  computation  of  the  internal  forces.  ] 


§  3.  Equilibrium  of  a  System  of  Bodies. 

119.  External  and  Internal  Forces  Acting  on  a  System  of 
Bodies. —  For  certain  purposes  it  is  found  convenient  to  group  a 
number  of  bodies  or  particles  together  in  applying  the  principles  of 
Statics,  such  a  group  being  called  a  system. 

When  a  system  of  bodies  is  considered,  a  force  acting  upon  any 
body  of  the  system  is  called  internal  or  external  according  as  the 
body  exerting  the  force  is  or  is  not  a  member  of  the  system. 

For  example,  every  terrestrial  body  is  attracted  by  the  earth  in 
accordance  with  the  law  of  gravitation.  This  attraction  is  external  if 
the  body  is  considered  by  itself.  If,  however,  the  body  and  the 
earth  are  regarded  as  forming  a  system,  the  force  mentioned  is  in- 
ternal. Thus,  whether  the  force  is  external  or  internal  depends  upon 
what  has  been  arbitrarily  chosen  as  the  system  to  be  discussed. 

120.  Conditions  of  Equilibrium  for  a  System  of  Bodies.  — It 

will  now  be  shown  that  if  every  body  in  a  system  is  in  equilibrium, 
the  external  forces  acting  on  the  system  as  a  whole  satisfy  the  same 
conditions  as  if  the  system  were  a  rigid  body  in  equilibrium. 

The  equations  of  equilibrium  may  be  written  for  each  body  of  the 
system  separately.  Let  a  resolution  equation  be  written  for  each 
body,  the  direction  of  resolution  being  the  same  for  all  ;  and  let  a 
moment  equation  be  written  for  each  body,  the  same  origin  of  mo- 
ments being  taken  for  all.  Adding  all  the  resolution  equations,  the 
resulting  equation  shows  that 

The  algebraic  sum  of  the  resolved  parts  of  all  forces  acting  upon 
the  bodies  is  equal  to  zero,  whatever  the  direction  of  resolution. 

Similarly,  the  addition  of  the  moment  equations  shows  that 

The  algebraic  sum  of  the  moments  of  all  the  forces  acting  upon 
the  bodies  is  equal  to  zero,  whatever  the  origin  of  moments. 

Now,  these  equations  include  both  external  and  internal  forces. 
But  since  the  forces  exerted  by  any  two  of  the  bodies  upon  each 


IOO  THEORETICAL    MECHANICS. 

other  are  equal  and  opposite  and  have  the  same  line  of  action,  the 
sum  of  the  resolved  parts  of  two  such  forces  in  any  direction  is  zero, 
and  the  sum  of  their  moments  about  any  origin  is  zero.  Hence  the 
sum  of  the  resolved  parts  of  all  internal  forces  in  any  direction  is 
zero,  and  the  sum  of  the  moments  of  all  internal  forces  about  any 
origin  is  zero.     The  following  proposition  may  therefore  be  stated  : 

If  every  body  of  a  system  is  in  equilibrium,  the  external  forces 
applied  to  the  system  satisfy  the  following  conditions  : 

(a)  The  sum  of  their  resolved  parts  in  any  direction  is  zero. 

(J?)   The  sum  of  their  moments  is  zero  for  any  origin* 

That  is,  the  external  forces  satisfy  the  same  conditions  as  if  the 
system  were  a  rigid  body. 

The  converse  of  this  proposition  is,  however,  not  generally  true. 
It  cannot  be  stated  that  when  the  external  forces  for  the  system  satisfy 
the  above  conditions,  the  forces  acting  on  each  separate  body  also 
satisfy  the  conditions  of  equilibrium. 

Although  the  general  principle  just  deduced  is  often  useful  in  the 
solution  of  problems,  a  complete  solution  usually  requires  in  addition 
the  employment  of  some  at  least  of  the  equations  of  equilibrium  for 
the  bodies  taken  separately. 

121.  Application. —  Two  homogeneous  smooth  cylinders  rest,  with 
axes  horizontal,  against  each  other  and  against  two  inclined  planes 
making  given  angles  with  the  horizontal.  Required  to  determine 
the  position  of  equilibrium. 

Regarding  the  two  cylinders  as  a  system,  the  external  forces  are 
the  weights  of  the  cylinders  (taken  as  acting  vertically  through  their 
centers),  and  the  pressures  exerted  by  the  supporting  planes  (acting 
normally  to  the  surfaces).  The  pressures  exerted  by  the  cylinders 
upon  each  other  are  internal  to  the  system. 

Two  solutions  will  be  given  ;  one  partly  geometrical,  the  other 
algebraic.     (See  Fig.  77.) 

Let  r'  and  r"  denote  the  radii  of  the  cylinders ;  W  and  W" 
their  weights  ;  6  the  angle  between  the  horizontal  and  the  plane 
containing  the  axes  of  the  cylinders  ;  a  and  /3  the  angles  made  by 
the  planes  with  the  horizontal ;  R '  and  R "  the  normal  pressures 
exerted  by  the  planes. 

(1)  Geometrical  solution. — Applying  the  conditions  of  equi- 
librium to  the  external  forces  acting  upon  the  system,  notice  that  the 
resultant  of  W  and  W"  is  a  vertical  force  of  magnitude  W  -f-  W" > 


EQUILIBRIUM    OF    A   SYSTEM    OF    BODIES. 


IOI 


and  is  in  equilibrium  with  R'  and  R",     Since  their  directions  are 
known,  their  magnitudes  may  be  determined  by  means  of  a  triangle. 


Fig.  77. 

Draw  LN  (Fig.  78)  equal  and  parallel  to  W  -f-  W'\  and  make 
NH  and  HL  parallel  respectively  to  R "  and  R ' ;  then  NH  repre- 
sents R"  in  magnitude  and  direction,  and 
HL  represents  R '.  Also,  the  angle  HLN 
=  a,  and  HNL  =  j3.     Hence 


R' 


R" 


sin  /3       sin  a 
or    R'  =  (W  -f   W7") 


FT'  +  W" 

sin  (a  -f-  /3) 

sin  /3 


sin  (a  +  £) 


sin  (a  -(-  /3) 

Next,  to  determine  0,  notice  that  the  re- 
sultant of  W  and  R'  acts  in  a  line  through 
A,  and  the  resultant  of  W"  and  7?"  acts  in 
a  line  through  i>.  Since  these  two  result- 
ants balance  each  other,  each  must  act  along 
the  line  AB.  Now  if,  in  Fig.  78,  the  point 
M  be  so  taken  that  LM  =  W  and  MN 
=  W'\  the  resultant  of  R '  and  W  is  rep- 
resented in  magnitude  and  direction  by  HM,  while  MH  represents 


102  THEORETICAL    MECHANICS. 

the  magnitude  and  direction  of  the  resultant  of  W"  and  R" . 
Hence,  for  equilibrium,  the  line  AB  must  be  parallel  to  HM.  In 
the  triangle  LMH  we  have  angle  at  L  =  a  ;  angle  at  H  —  90°  — 
(a  +  0)  ;  angle  at  J/  =  90°  -f  0  ; 

W  R' 


hence 


or 


sin  (90°  —  a  —  0)       sin  (90°  -J"  0) 
W  R' 


cos  (a  -f-  0)       cos  0 
Solving  for  0, 

cos  (a  +  0)  _     W  _  FT'  sin  (a  +  g) 

cos0  "  tf'  "~   JF'  +   IF*  '         sin  /3        ' 

r     „                     „         W7"  cotan  a  —   W  cotan  /3 
or,  finally,  tan  a  — . 

(2)  Algebraic  solution. — Applying   the   algebraic  conditions  of 
equilibrium  to  the  four  forces  W,   W" ,  R'  and  R"}  three  inde- 
pendent equations  may  be  written  as  follows  : 
Resolving  horizontally, 

R'  sin  a  — R"  sin  ft  =  o.       .        .        .     (1) 
Resolving  vertically, 

R'  cos  a  +  R"  cos  0  —  (  W  +   W")  =  o.         .     (2) 

Taking  moments  about   .5,  and  noticing   that   the  arm   of  R'  is 
(r'  -f  r")cos(a  +  0), 

^'  (r*  +  r")  cosO  —  R'  (r'  +  r^  cos  (a  +  0)  =  o.      (3) 

These  three  equations  contain  three  unknown  quantities,  R',  R" 
and  0.  From  equations  (1)  and  (2),  R'  and  R"  maybe  found ;  and 
then  0  can  be  determined  from  (3).  The  results  will  agree  with 
those  found  above. 

If  it  is  desired  to  determine  the  pressure  between  the  two  cylin- 
ders, the  conditions  of  equilibrium  must  be  applied  to  one  of  the 
cylinders  alone. 

§  4.    Stress. 

122.  External  and  Internal  Stresses. —  A  stress  has  already 
been  defined  as  consisting  of  two  equal  and  opposite  forces  exerted 
by  two  portions  of  matter  upon  each  other  (Art.  36).  The  two 
forces  of  a  stress  always  constitute  an  ' '  action  ' '  and  its  ' '  reaction. ' ' 


STRESS.  IO3 

When  any  body  or  system  of  bodies  is  under  consideration,  the 
stresses  with  which  the  system  is  concerned  may  be  either  external 
or  internal.  An  external  stress  is  one  exerted  between  two  portions 
of  matter,  one  of  which  belongs  to  the  body  or  system  while  the 
other  does  not.  An  internal  stress  is  one  exerted  between  two  por- 
tions of  matter  both  of  which  belong  to  the  body  or  system. 

This  classification  expresses  no  distinction  as  to  the  nature  of  the 
stresses.  It  is  merely  a  distinction  made  for  convenience  in  dealing 
with  the  problems  of  Mechanics. 

123.  Kinds  of  Internal  Stress  in  a  Body. —  Considering  any 
two  adjacent  portions  of  a  body,  separated  by  a  plane  surface,  the 
resultant  stress  between  these  two  portions  may  have  any  direction. 
Whatever  this  direction  may  be,  let  each  of  the  forces  of  the  stress  be 
resolved  into  two  components,  one  normal  to  the  surface  of  separa- 
tion and  the  other  parallel  to  it.  The  normal  components  of  the 
forces  constitute  a  normal  stress,  and  the  other  components  a  tan- 
gential stress. 

The  normal  stress  may  be  either  tensile  or  compressive.  The 
former  resists  a  tendency  of  the  two  portions  of  the  body  to  separate ; 
the  latter  resists  a  tendency  of  the  two  portions  to  move  toward  each 
other. 

The  tangential  stress  is  called  a  shearing  stress.  It  resists  a  tend- 
ency of  the  two  portions  of  the  body  to  slide  past  each  other  along 
the  surface  separating  them. 

To  illustrate  the  two  kinds  of  normal  stress,  reference  may  be 
made  to  the  problem  discussed  in  Art.  118.  The  roof  truss  being 
regarded  as  consisting  of  two  parts,  X  and  K,  it  was  found  that  Y 
exerts  upon  X  three  forces  whose  magnitudes  and  directions  are 
shown  in  Fig.  7  5  ;  while  X  exerts  upon  Y  forces  equal  and  opposite 
to  these.  It  is  thus  seen  that  normal  stresses  act  in  every  trans- 
verse section  of  each  bar ;  the  stress  in  DC  being  compressive,  while 
those  in  BC  and  BE  are  tensile  stresses.  The  nature  of  the  normal 
stress  in  any  bar  of  a  jointed  frame  may  thus  readily  be  found  by  de- 
termining the  direction  of  the  internal  force  exerted  by  one  part  of 
the  structure  upon  the  other. 

Examples. 

1.  Determine  the  kind  of  stress  in  each  of  the  bars  of  the  truss 
shown  in  Fig.  73,  due  to  the  loads  described  in  Ex.  1,  Art.  118. 


104  THEORETICAL    MECHANICS. 

2.  In  Ex.  2,  Art.  118,  determine  the  kind  of  stress  in  each 
member  of  the  truss. 

124.  Kinds  of  Stress  Between  Bodies. —  If  two  bodies  are  in 
contact,  the  forces  which  they  exert  upon  each  other  at  the  surface 
of  contact  constitute  stresses  which  may  be  classified  in  the  same 
manner  as  the  internal  stresses  in  a  body.  The  names  tension,  com- 
pression and  shear  are,  however,  confined  to  the  internal  stresses. 

The  normal  stress  between  two  bodies  in  contact  is  usually  pres- 
sure;  that  is,  the  forces  have  such  directions  as  to  resist  a  tendency 
of  the  bodies  to  move  toward  each  other.  If  the  bodies  for  any 
reason  have  a  tendency  to  separate,  they  are  in  general  able  to  exert 
only  very  slight  forces  to  resist  this  tendency. 

The  tangential  stress  between  two  bodies  at  their  surface  of  con- 
tact can  have  any  magnitude  up  to  a  certain  limit.  Thus,  if  a  body 
rests  upon  a  horizontal  surface,  and  if  a  horizontal  force  is  applied  to 
it,  the  supporting  body  exerts  a  force  in  the  opposite  direction.  If 
the  applied  force  does  not  become  too  great,  the  opposing  force  will 
always  just  equal  it  and  the  body  will  remain  in  equilibrium.  A  tan- 
gentialstress thus  acts  between  the  two  bodies.  The  forces  compos- 
ing such  a  stress  are  of  the  kind  to  which  the  name  friction  is  ap- 
plied.    The  subject  of  friction  is  treated  in  Chapter  VII. 

Besides  the  stresses  acting  between  bodies  at  their  surfaces  of  con- 
tact, there  are  other  stresses  which  act  between  bodies  not  in  contact 
and  without  any  apparent  material  connection.  These  are  the  so- 
called  *  ■  actions  at  a  distance ' '  already  mentioned  (Art.  40).  Such 
stresses  are  frequently  called  attractions  and  repulsions,  and  are  anal- 
ogous to  tensile  and  compressive  stresses  respectively. 

125.  Strain. —  The  external  forces  applied  to  a  body  usually  tend 
to  cause  the  different  parts  of  the  body  to  move  relatively  to  each 
other.  This  tendency  is  opposed  by  the  internal  forces  called  into 
action.  No  natural  body,  however,  remains  wholly  rigid  under  ap- 
plied forces,  but  the  parts  move,  at  least  slightly,  relatively  to  one 
another. 

Strain  is  the  name  given  to  the  deformation  which  a  body  under- 
goes under  the  action  of  applied  forces. 

In  this  work  we  shall  not  usually  be  concerned  with  strain,  the 
bodies  considered  being  regarded  as  rigid.  Any  ordinary  solid  body 
will  assume  a  form  of  equilibrium  under  the  applied  forces  (if  they 


STRESS.  IO5 

are  not  such  as  to  cause  rupture)  and  may  then  be  regarded  as  a 
strictly  rigid  body. 

Examples. 

1.  A  straight  bar  of  uniform  cross-section  and  density,  14  ft.  long 
and  weighing  1 20  lbs. ,  rests  in  a  horizontal  position  upon  smooth 
supports  at  the  ends.  Conceive  the  bar  to  be  divided  by  a  plane 
perpendicular  to  the  length,  4  ft.  from  one  end.  Determine  (a)  the 
resultant  tangential  stress  and  (b)  the  resultant  normal  stress  between 
the  two  portions  at  the  surface  of  separation. 

2.  With  the  data  of  Ex.  1,  take  a  transverse  section  through  the 
middle  of  the  bar,  and  determine  the  resultant  tangential  stress  and 
resultant  normal  stress  between  the  two  portions. 

3.  A  body  weighing  50  lbs.  is  at  rest  upon  a  rough  horizontal 
surface  while  acted  upon  by  a  force  of  10  lbs.  directed  at  an  angle  of 
300  upward  from  the  horizontal.  Describe  completely  the  resultant 
stress  acting  between  the  given  body  and  the  supporting  body. 

4.  A  uniform  bar  AB,  12  ft.  long,  weighing  18  lbs.,  is  supported 
at  A  by  a  smooth  hinge  and  at  B  by  a  string  inclined  30°  to  the 
vertical.  Conceiving  the  bar  divided  in  the  middle  by  a  transverse 
plane,  determine  (a)  the  resultant  normal  stress,  (b)  the  resultant 
tangential  stress,  and  (V)  the  resultant  stress,  acting  between  the  two 
portions  of  the  body. 


CHAPTER  VII. 

FRICTION. 

126.  Smooth  and  Rough  Surfaces. —  The  conception  of  a  per- 
fectly smooth  surface  has  frequently  been  employed  in  the  foregoing 
discussions.  Such  a  surface  was  defined  in  Art.  42.  The  surfaces  of 
actual  bodies  are,  however,  always  more  or  less  rough. 

If  any  two  bodies  are  in  contact,  each  exerts  upon  the  other  a 
force,  the  direction  of  which  depends  upon  various  conditions.  Let 
the  surface  of  one  of  the  bodies  be  a  plane,  and  consider  the  force  P 
which  this  body  exerts  upon  the  other.  Whatever  be  the  direction 
of  this  force,  let  it  be  resolved  into  two  components  N  and  T,  the 
former  perpendicular  to  the  plane  surface  and  the  latter  parallel  to  it. 
(Fig.  79  shows  P%  N  and  T  in  magnitude  and  direction.  The  lines 
of  action  of  P  and  N  are  also  shown  ;  but  T  will 
act  along  a  line  lying  in  the  surface  of  contact. ) 
If  the  plane  surface  be  very  rough,  the  magnitude 
of  T  may  be  considerable  in  comparison  with  N; 
but  if  the  surface  be  made  smoother,  the  magni- 
tude of  the  tangential  force  which  can  be  exerted 
becomes  smaller.  We  are  thus  led,  as  in  Art. 
Fig.  79.  42,  to  define  a  perfectly  smooth  surface  as  one 

which  can  exert  no  force  parallel  to  itself  upon 
any  body  placed  in  contact  with  it.  In  other  words,  the  resultant 
pressure  exerted  by  a  smooth  surface  upon  any  body  in  contact  with 
it  must  have  the  direction  of  the  normal  to  the  surface. 

Although  the  surface  of  a  body  can  never  be  made  to  satisfy  this 
definition  of  perfect  smoothness,  there  are  cases  which  approach  near 
to  it.  The  surface  of  a  sheet  of  ice  is  often  nearly  smooth,  and  the 
difficulty  of  walking  on  such  a  sheet  is  due  to  the  fact  that  only  a 
small  tangential  force  can  be  exerted  by  the  ice  upon  a  body  in  con- 
tact with  it. 

The  forces  exerted  by  two  bodies  upon  each  other  at  their  surface 
of  contact  are  of  the  kind  called  passive  resistances  (Art.  41).  They 
are  called  into  action  to  resist  a  tendency  of  the  two  bodies,  due  to 
any  cause,  to  move  relatively  to  each  other.     Thus,  suppose  a  body 


FRICTION.  107 


to  rest  upon  a  horizontal  surface  and  to  be  acted  upon  by  the  attrac- 
tion of  the  earth ;  i.  c,  by  a  downward  force  equal  to  the  weight  of 
the  body.  The  pressure  exerted  upon  it  by  the  table  is  called  into 
action  to  resist  this  downward  force  and  is  exactly  equal  and  opposite 
to  it.  Let  a  horizontal  force  be  now  applied  to  the  body.  If  this 
force  be  not  too  great  the  body  will  remain  at  rest,  and  in  order  to 
hold  it  at  rest  the  supporting  body  exerts  upon  it,  in  addition  to  the 
vertical  force  already  mentioned,  a  force  equal  and  opposite  to  the 
horizontal  applied  force.  If  the  magnitude  of  the  horizontal  force  be 
gradually  changed  from  zero  to  any  value  (up  to  a  certain  limit),  or 
if  its  direction  be  changed,  the  resisting  force  changes  in  a  cor- 
responding manner  so  as  always  to  be  equal  and  opposite  to  it.  The 
resultant  force  exerted  by  the  supporting  body  is  made  up  of  the 
upward  force  resisting  the  weight  of  the  supported  body,  and  the 
horizontal  force  just  described.  If  additional  forces  be  applied  to 
the  body,  equal  and  opposite  forces  will  be  exerted  by  the  supporting 
body ;  but  the  magnitude  and  direction  of  the  force  that  can  be 
resisted  are  always  subject  to  limitation,  depending  upon  the  material 
composing  the  bodies  and  the  nature  of  their  surfaces. 

127.  Friction. —  The  name  friction  is  given  to  the  tangential 
component  of  the  force  exerted  by  one  body  upon  another  at  their 
surface  of  contact.  The  frictional  forces  exerted  by  the  two  bodies 
upon  each  other  are  equal  and  opposite  and  constitute  a  stress. 

Experiments  show  that  friction  acts  according  to  certain  laws. 
In  stating  these  laws,  the  term  "applied  forces"  will  be  used  to  desig- 
nate all  forces  acting  upon  the  body  in  question,  except  the  "passive 
resistance"  exerted  by  the  other  body  at  their  surface  of  contact. 

128.  Laws  of  Friction.  —  The  following  may  be  stated  as  the 
laws  or  principles  to  be  employed  in  discussing  the  force  of  friction 
exerted  upon  a  body  by  another  in  contact  with  it. 

(1)  If  the  body  is  in  equilibrium  the  friction  is  equal  and  oppo- 
site to  the  tangential  component  of  the  resultant  of  the  applied  forces. 

(2)  If  the  tangential  component  of  the  resultant  of  the  applied 
forces  becomes  greater  than  a  certain  limiting  value,  the  friction  can- 
not become  great  enough  to  balance  it.  The  value  of  the  friction 
when  sliding  is  about  to  take  place  is  called  the  limiting  friction. 

(3)  The  magnitude  of  the  limiting  friction  bears  a  constant  ratio 
to  that  of  the  normal  pressure  between  the  two  bodies. 

(4)  The  ratio  of  the  limiting  friction  to  the  normal  pressure  is 


108  THEORETICAL    MECHANICS. 

independent  of  the  area  of  contact  of  the  two  bodies,  if  the  touching 
surfaces  are  uniform  in  character. 

(5)  If  the  body  is  sliding  along  the  surface  of  contact  the  friction 
is  independent  of  the  velocity  and  proportional  to  the  normal  pres- 
sure. In  this  case,  the  ratio  of  the  friction  to  the  normal  pressure  is 
found  to  be  less  than  when  sliding  is  about  to  occur. 

Of  these  five  laws,  (1)  and  (2)  are  exact.  The  first  is,  indeed,  a 
direct  result  of  the  principles  of  equilibrium.  Laws  (3),  (4)  and  (5) 
are  only  approximate,  but  are  nearly  enough  true  for  most  practical 
applications. 

129.  Coefficient  of  Friction. —  The  ratio  of  the  magnitude  of 
the  limiting  friction  to  that  of  the  normal  pressure  between  the  bodies 
is  called  the  coefficient  of  friction.     That  is,  if  jjl  is  the  coefficient  of 

friction, 

fi  =  FIN, 

in  which  F  denotes  the  limiting  friction  and  N  the  normal  pressure. 

Law  (3)  states  that  //,  is  constant  for  two  given  bodies,  whatever 
the  magnitude  of  N.  If  the  pressure  between  the  two  bodies  be- 
comes nearly  great  enough  to  crush  the  material  of  which  either  is 
composed,  it  is  found  that  /jl  is  considerably  greater  than  for  small 
pressures.  Within  certain  limits  of  normal  pressure,  however,  its 
value  is  nearly  constant,  and  it  will  be  regarded  as  constant  in  the 
applications  which  follow. 

It  is  to  be  noted  that  the  value  of  /x  depends  upon  the  nature  of 
both  the  bodies  concerned. 

130.  Angle  of  Friction. — The  total  pressure  exerted  by  either 
body  upon  the  other  is  the  resultant  of  the  normal  pressure  and  the 
friction.  If  this  resultant  be  determined  by  the  triangle  of  forces  it 
is  seen  that  its  direction  makes  with  the  normal  an  angle  whose  tan- 
gent is  the  ratio  of  the  friction  to  the  normal  pressure.  This  angle 
has  its  greatest  value  when  the  friction  is  as  great  as  possible. 

The  angle  of  friction  is  the  angle  between  the  resultant  pressure 
and  the  normal  when  sliding  is  about  to  take  place. 

If  this  angle  is  denoted  by  </>,  it  follows  from  the  definition  that 

tan  (j>  =  F/N=  /*. 

That  is,  the  tangent  of  the  angle  of  friction  is  equal  to  the  coefficient 
of  friction. 


FRICTION. 


IO9 


131.  Experimental  Determination  of  Coefficient  of  Friction. — 

The  value  of  the  coefficient  of  friction  may  often  be  determined  very 
simply.  Thus,  let  a  body  be  placed  upon  a  horizontal  plane  surface 
and  let  a  force  be  applied  horizontally  just  great  enough  so  that  the 
body  is  on  the  point  of  sliding.  This  force  is  equal  and  opposite  to 
the  limiting  friction.  Its  magnitude  may  be  determined  by  a  spring 
balance.  The  normal  pressure  is  in  this  case  equal  to  the  weight  of 
the  body,  which  may  also  be  determined  by  the  spring  balance.  The 
ratio  of  these  two  forces  is  the  coefficient  of  friction. 

Another  method  of  determining  the  coefficient  of  friction  is  as 
follows :  Let  a  body  be  placed  upon  an  inclined  plane,  and  let  no 
forces  act  upon  it  except  its  weight  and  the  pressure  of  the  supporting 
plane.  This  supporting  pressure  is  exactly  equal  and  opposite  to 
the  weight  so  long  as  the  body  is  in  equilibrium  ;  the  angle  it  makes 
with  the  normal  is  therefore  equal  to  the  inclination  of  the  plane  to 
the  horizontal.  Let  this  inclination  be  gradually  increased  until  the 
body  is  on  the  point  of  sliding.  The  resultant  pressure  exerted  by 
the  plane  now  makes  with  the  normal  an  angle  equal  to  the  angle  of 
friction.  Hence,  if  the  inclination  of  the  plane  in  this  limiting  posi- 
tion be  measured,  the  coefficient  of  friction  can  be  found  from  the 
relation 

fjL  =  tan  <f>. 

Many  experiments  have  been  made  to  determine  the  value  of  the 
coefficient  of  friction  for  different  materials  both  with  and  without 
lubricants.  The  range  of  the  results  is  indicated  by  the  following 
values  *  of  /x : 


Wood  on  wood,  dry 

. 

0.25  to  0.5 

"      "      "     soaped 

. 

0.2 

Metals  on  oak,  dry 

. 

2.5    to  0.6 

m      "      wet 

. 

0.24  to  0.26 

"       "      "      soaped 

0.2 

Metals  on  metals,  dry 

0.15  to  0.2 

<<            i<             a             wet 

0.3 

Smooth  surfaces  occasionally  1 

ubricated 

0.07  to  0.08 

"              V        thoroughly 

t  * 

0.03  to  0.036 

Ex 

AMPLES. 

1.  A  body  of  lV\bs.  mass  is  at  rest  upon  a  plane  making  angle 
0  with  the  horizontal.  A  cord  attached  to  this  body  runs  parallel  to 
the  plane,  passes  over  a  smooth  pulley  and  sustains  a  weight  of  P 


See  Encyclopaedia  Britannica,  Vol.  XV,  p.  765. 


IIO  THEORETICAL    MECHANICS. 

lbs.  Determine  the  magnitude  and  direction  of  the  friction,  the 
normal  pressure,  and  the  total  pressure,  exerted  by  the  plane  upon 
the  body. 

2.  In  Ex.  i,  let  W  =  50  lbs.,  P  =  40  lbs.,  0  =  320,  and  sup- 
pose the  body  is  just  about  to  slide  up  the  plane.  Determine  the 
coefficient  of  friction.  Ans.  ^  =  0.318  ;  <£=  170  40'. 

3.  In  Ex.  1,  if  the  coefficient  of  friction  is  0.4  and  0  is  300,  what 
must  be  the  value  of  P  (in  terms  of  W)  in  order  that  the  body  may 
just  slide  up  the  plane?  What  value  of  P  will  just  allow  sliding 
down  the  plane?  Ans.  0.846  W ';  0.154  IV. 

4.  A  body  of  30  lbs.  mass,  resting  upon  a  plane  inclined  450  to 
the  horizontal,  is  pulled  horizontally  by  a  force  P.  If  the  coefficient 
of  friction  is  o.  2,  between  what  limits  may  the  value  of  P  vary  and 
still  permit  the  body  to  remain  at  rest?     Ans.  20  lbs.  and  45  lbs. 

5.  A  straight  bar  of  length  /,  whose  center  of  gravity  is  distant  a 
from  the  lower  end,  rests  upon  a  horizontal  surface  and  leans  against 
a  vertical  wall.  If  the  coefficient  of  friction  between  the  floor  and 
bar  is  /x,  and  the  wall  is  perfectly  smooth,  what  angle  with  the  verti- 
cal may  the  bar  make  without  sliding?  Ans.  tan-1  (fil/a). 

6.  If  the  conditions  are  as  in  Ex.  5,  except  that  the  vertical  wall 
is  rough,  the  coefficient  of  friction  between  the  wall  and  the  bar 
being  /x',  what  is  the  position  of  incipient  sliding  ? 

7.  What  are  the  limiting  positions  of  equilibrium  for  a  heavy 
bead  on  a  circular  wire  ring  whose  plane  is  vertical  ?  (  Express  the 
result  in  terms  of  the  coefficient  of  friction.) 

8.  What  are  the  limiting  positions  of  equilibrium  of  a  uniform 
bar  placed  wholly  within  a  spherical  bowl  ?  (  Express  in  terms  of 
the  angle  of  friction.) 

Ans.  Let  2a  =  angle  subtended  at  center  by  bar,  6  =  its 
angle  with  the  horizontal  in  the  limiting  position  ;  then  tan  6  = 
%  [tan  (a  -j-  cj>)  —  tan  (a  —  <£)]. 

9.  A  bar  A  B  whose  mass  is  6  lbs.  and  whose  center  of  gravity  is 
4  ft.  from  the  end  A  is  supported  in  a  horizontal  position  by  a 
string  attached  at  A  and  a  peg  7  ft.  from  A.  The  coefficient  of  fric- 
tion between  the  bar  and  the  peg  is  0.3.  If  the  bar  is  about  to  slide 
in  the  direction  AB,  determine  the  direction  of  the  string,  the  ten- 
sion it  sustains,  and  the  supporting  pressure  exerted  by  the  peg. 

Ans.  Tension  =  2.77  lbs.;  resistance  of  peg  =  3.58  lbs.;  angle 
between  string  and  bar  =  68°  12'. 

10.  In  the  preceding  case,  between  what  limits  must  the  direction 
of  the  string  lie  in  order  that  there  may  be  equilibrium  ?  Determine 
the  corresponding  limiting  values  of  the  supporting  forces. 

n.  Solve  the  preceding  problem  in  the  following  general  case: 
Let  W  =  weight  of  bar,  a  =  distance  of  its  center  of  gravity  from 
A,  b  =  distance  of  peg  from  A,  fi  =  coefficient  of  friction. 


FRICTION. 


Ill 


132.  Friction  in  Machines.— In  order  that  a  simple  machine 
may  produce  its  desired  effect,  its  motion  must  be  constrained,  that 
is,  guided  in  a  definite  manner.  This  is  accomplished  by  means  of 
bodies  which  touch  the  machine  parts  at  certain  points  and  exert 
constraining  forces  upon  them.  Besides  the  forces  which  produce 
the  desired  constraint,  there  are  usually  frictional  forces  which  resist 
any  tendency  of  the  bodies  to  slide  over  each  other  at  their  surfaces 
of  contact.  In  general  friction  is  a  wasteful  resistance ;  but,  as  will 
be  seen,  it  may  in  certain  cases  act  with  the  effort  and  thus  aid  the 
useful  object  of  the  machine. 

The  effect  of  friction  on  the  operation  of  a  machine  may  be 
estimated  by  means  of  the  laws  of  friction  above  stated.  The 
method  of  applying  them  is  illustrated  in  the  following  simple 
case. 

133.  Fixed  Pulley  with  Friction.— If  a  pulley  is  mounted 
freely  upon  a  cylindrical  axle,  the  frictional  force  exerted  by  the 
axle  upon  the  wheel  always  acts  in  such  a  way  as  to  oppose  what- 
ever rotation  the  wheel  has  or  tends 

to  have  by  reason  of  the  action   of 
other  forces. 

Referring  to  Fig.  80,  let  a  de- 
note the  radius  of  the  pulley;  r  the 
radius  of  the  axle  ;  P  the  effort ;  W 
the  load  ;  R  the  resultant  pressure 
exerted  by  the  axle  on  the  pulley  ; 
N  the  normal  component  of  R ;  F 
the  friction,  or  tangential  component 
of  R  ;  fi  the  coefficient  of  friction  ; 
cf)  the  angle  of  friction. 

The  cylindrical  hole  into  which 
the  axle  fits  is  slightly  larger  than  the 
axle,   but  they  may  be  regarded   as 

practically  equal  in  size.  If  the  wheel  is  on  the  point  of  moving 
with  the  effort,  the  equation  of  moments,  taking  origin  at  center 
of  wheel,  is 

Pa  —  Wa  —  Fr  =  o, 


Fig.  80. 


or 


P=  W+Fr/a, 


(1) 


which  shows  a  mechanical  disadvantage. 


112  THEORETICAL    MECHANICS. 

If  the  wheel  is  on  the  point  of  moving  against  the  effort,  the 
direction  of  F  is  reversed,  and  the  equation  is 

P  =  W—Fr/a,      ....      (2) 

showing  a  mechanical  advantage. 

The  value  of  F  in  case  of  incipient  motion  may  be  determined  as 
follows: 

In  this  limiting  case, 

F  =  N  tan  </>, 

the  total  resistance  R  acting  at  an  angle  (j>  with  the  normal  to  the 
surfaces  at  the  point  of  contact.  But  this  total  resistance  must  act 
in  a  line  through  the  intersection  of  the  lines  of  action  of  P  and  W. 
Let  0  be  the  value  of  the  angle  CAB  (Fig.  80);  then 

sin  0  sin  <f> 

r  a  cosec  a 

. '.     sin  0  =1  -  sin  <f>  sin  a.  (3) 

a 

Since  R,  P  and  W  are  in  equilibrium,  we  have 
R  P  W 


sin  2a       sin  (a  -f-  0)        sin  (a  —  0) 

D  P  sin  2a  W  s\n  2<x 

K  = =  ;       .  .      (4) 

sin  (a  +  0)        sin  (a  —  0) 

_        „    .      ,         W  sin  2a  sin  6  ,  _ 

F  =  tf  sin  £  =  —— — X.        .  .      (5) 

sin  (a  —  0) 

From  equation  (5)  the  value  of  F  in  terms  of  W  can  be  found  after 
0  is  determined  from  equation  (3).  Substituting  the  value  of  F  in 
equation  (1),  the  relation  between  Pand  W  is  determined. 

The  above  discussion  applies  to  the  case  in  which  there  is  incipi- 
ent motion  with  the  force  P.  If  the  opposite  motion  is  about  to 
occur,  the  form  of  the  result  is  similar,  but  P  and  W  must  be  inter- 
changed. 

Examples. 

1.  Let  /^and  W  (Fig.  80)  be  inclined  to  each  other  at  an  angle 
of  900;  radius  of  pulley  =  6  ins.;  radius  of  axle  =  ^  in. ;  coefficient 
of  friction  ==  o.  2.  Determine  0  and  the  relation  between  P  and  W 
in  case  of  incipient  motion  in  each  direction. 

Ans-  0  =  59';  PjW  =  0.952  or  1.050. 


FRICTION.  113 

2.  Prove  that,  if  P  and  W  are  parallel,  the  relation  between 
them  is  pjW  —  (a  -f  r  sin  cf>)  /  (a  —  r  sin  <£) 

or  P/lV=(a  —  r  sin  <j>)/(a  -\-  r  sin  $), 

according  as  there  is  incipient  motion  with  P  or  against  P.  Show 
that  the  first  of  these  equations  is  included  as  a  special  case  in  the 
above  general  solution. 

3.  With  data  as  in  Ex.  1 ,  except  that  P  and  W  are  parallel,  de- 
termine the  relation  between  P  and  W  iox  both  cases  of  incipient 
motion.      Determine  also  the  value  of  /^and  of  R  in  each  case. 

Ans.  Pj  W '=  0.952  or  1.050. 
8 


CHAPTER    VIII. 

EQUILIBRIUM    OF    FLEXIBLE    CORDS. 

134.  Natural  Bodies  Not  Perfectly  Rigid. —  A  perfectly  rigid 
body  could  not  be  deformed  in  any  degree  by  the  action  of  external 
forces.  Actual  bodies,  however,  undergo  changes  of  form  and  size 
when  forces  are  applied  to  them.  In  some  cases  the  changes  are  so 
slight  as  to  be  of  little  importance.  Thus,  a  steel  beam,  resting  upon 
two  supports,  may  bend  only  slightly  under  very  heavy  loads.  For 
certain  purposes  this  bending  is  unimportant,  while  for  others  it  must 
be  taken  into  account.  The  statics  of  non-rigid  bodies  is  less  simple 
than  that  of  rigid  bodies.  Certain  classes  of  problems  may,  how- 
ever, be  treated  without  great  difficulty  by  means  of  the  following 
general  principle. 

135.  Non-Rigid  Body  in  Equilibrium.  — The  conditions  of  equi- 
librium for  a  rigid  body  apply  to  any  body,  every  portion  of  which 
is  in  equilibrium. 

Suppose  a  body,  acted  upon  by  any  external  forces,  to  be  de- 
formed in  any  manner,  reaching  finally  a  condition  of  equilibrium. 
This  condition  being  attained,  the  rigidity  of  the  body  is  of  no  further 
consequence  so  far  as  its  equilibrium  is  concerned.  The  external 
forces  must  therefore  satisfy  the  same  conditions  as  if  the  body  were 
actually  rigid. 

136.  Flexible  and  Inextensible  Cord. —  A  flexible  cord  has 
been  defined  as  one  which  may  be  bent  by  the  application  of  exter- 
nal forces.  A  perfectly  flexible  cord  would  offer  no  resistance  to 
bending. 

The  following  discussions  will  relate  to  cords  assumed  to  possess 
this  ideal  property  of  perfect  flexibility.  The  results  are  of  practical 
value,  since  many  actual  cords  are  flexible  to  a  high  degree.  It  will 
be  assumed  further  that  the  cords  are  inextensible. 

The  forces  acting  upon  a  cord  may  be  concentrated  or  distributed. 
(Art.  37).     These  two  cases  will  be  considered  separately. 

137.  Geometrical  Conditions  of  Equilibrium  of  Cord  Acted 
Upon  by  Concentrated  Forces. —  Consider  a  perfectly  flexible, 
weightless  cord,  suspended  from  two  fixed  points,  and  acted  upon  by 


EQUILIBRIUM    OF    FLEXIBLE    CORDS. 


"5 


any  concentrated  forces.  Thus,  let  Jf  and  Y  (Fig.  81)  be  the  fixed 
points  of  suspension,  and  let  forces  Px,  P2,  P3  be  applied  at  points  L, 
M,  N.  Suppose  the  cord  to  have  assumed  a  form  of  equilibrium,  the 
segments  XL,  LM,  MN,  NY  being  straight.  Let  Tx,  T2i  T3,  TA 
denote  the  tensions  in  the  successive  segments  of  the  string,  taken  in 
order  from  X  to  Y.  Then  there  is  a  definite  relation  between  any- 
applied  force,  as  Px ,  and  the  tensions  in  the  adjacent  portions  of  the 
string,  as  Tx  and  T2 .  By  Art.  115  any  portion  of  the  string  may  be 
treated  as  a  separate  body,  and  the  conditions  of  equilibrium  may 
be  applied  to  the  forces  acting  upon  it.  For  the  portion  of  the 
string  near  L  the  forces  are  three,  namely,  Px  and  the  forces  due  to 
the  tensions  in  XL  and  LM, —  that  is,  a  force  Tx  in  the  direction 
LX,  and  a  force  T%  in  the  direction  LM.     Similarly,  for  the  por- 


Fig.  81. 


tion  of  the  string  near  M  we  have  three  forces  in  equilibrium :  the 
force  Ptt  a  force  T2  in  the  direction  ML,  and  a  force  Ts  in  the 
direction  MN.  A  similar  analysis  applies  to  the  point  N.  If  Plt 
P2  and  Pz  are  known,  the  tensions  may  be  found,  either  geometri- 
cally or  by  writing  the  equations  of  equilibrium  for  each  of  the  sets 
of  forces  named.  The  geometrical  construction  is  shown  in  Fig.  81. 
The  line  AB  is  drawn  to  represent  Px  in  magnitude  and  direc- 
tion ;  then  BO  and  OA  drawn  parallel  to  LM  and  LX  represent  the 
forces  T.2  and  Tx.  Draw  BC  to  represent  P2  in  magnitude  and  direc- 
tion ;  then  for  the  equilibrium  of  the  portion  of  the  cord  near  M, 
CO  must  1  epresent  the  force  T.A .  Similarly,  if  CD  represents  P3  in 
magnitude   and   direction,  DO  must   represent  the  force  TK.    The 


Il6  THEORETICAL    MECHANICS. 

figure  *  thus  shows  at  a  glance  the  relations  of  magnitude  and  direc- 
tion that  must  subsist  among  the  forces  Ply  P3i  P.6  and  the  tensions 

-*  1  J     *  %  I     -*  3  >     M  4  • 

Algebraically,  the  relations  between  these  quantities  may  be  ex- 
pressed by  writing  two  equations  of  equilibrium  for  each  of  the  three 
systems  of  forces  acting  at  L,  M  and  N. 

138.  General  Problem  of  Cord  Acted  Upon  by  Concentrated 
Forces. —  If  a  cord  fixed  at  two  points  is  acted  upon  by  concentrated 
forces,  numerous  problems  may  arise,  depending  upon  what  quanti- 
ties are  given  and  what  are  required.  Thus,  in  the  case  represented 
in  Fig.  8 1 ,  the  quantities  involved  are  the  applied  forces  Px ,  P.l  and 
PA ;  the  tensions  Tx ,  T.t ,  T3 ,  Tt ;  and  the  coordinates  which  give 
the  positions  of  the  points  X,  L,  M,  N,  Y.  (Instead  of  these  coordi- 
nates, we  may  use  the  lengths  and  directions  of  the  segments  XL, 
LM,  etc.)  If  enough  of  these  quantities  are  known  the  others  may 
be  determined.  Many  of  the  problems  which  may  arise  do  not  admit 
of  simple  treatment.  If  the  problem  is  determinate,  as  many  equa- 
tions may  be  written  as  there  are  unknown  quantities.  Part  of  these 
are  the  equations  of  equilibrium,  written  for  each  point  at  which  a 
force  is  applied  in  the  manner  above  indicated  ;  the  others  must  be 
written  from  the  geometrical  relations  of  the  figure. 

The  following  examples  illustrate  simple  cases. 

Examples. 

1.  Let  a  single  vertical  load  of  /Mbs.  be  suspended  from  a  cord 
at  distances  /  and  m  from  its  ends  ;  and  let  the  ends  X  and  Y  be 
fixed  in  such  positions  that  the  distance  XY  is  a  and  the  line  XV 
makes  with  the  horizontal  an  angle  6.  Determine  the  tensions  in 
the  cord. 

2.  In  the  preceding  example  let  /  =■  10  ins.,  m  =  8  ins.,  a  — 
12  ins.,  6  =  300.  Ans.  0.0744  P  and  0.988  P. 

3.  Let  the  ends  of  the  cord  be  fixed  at  points  in  the  same  hori- 
zontal line  at  a  distance  apart  equal  to  a,  and  let  vertical  loads  /^and 
Q  be  suspended  at  points  dividing  the  length  of  the  cord  into  seg- 
ments /,  m  and  n.     Determine  the  relation  between  P  and  Q  so  that 

*  The  construction  here  described  is  of  importance  in  Graphical  .Statics. 
The  polygon  formed  by  the  successive  segments  of  the  cord  is  called  a  funic- 
ular polygon.  (Sometimes  called  also  equilibrium  polygon  and  string 
polygon.) 


EQUILIBRIUM    OF    FLEXIBLE    CORDS. 


117 


the  middle  segment  may  be  horizontal.  Prove  that,  if  the  middle 
segment  is  horizontal,  the  following  relation  must  be  satisfied : 
(P-  Q)I(P  +  Q)  =  («a  -  ni(a  -  my. 

4.  If,  in  the  preceding  example,  a  =  12  ins.  ,7=4  ins. ,  m  =  6 
ins.,  ?i  =  6  ins.,  prove  that  P  =  3.5Q.  Determine  the  tensions, 
assuming  Q  =  10  lbs.,  P  =  35  lbs. 

Ans.  Tension  in  horizontal  cord  ==  12.37  lbs. 

139.  Equations  of  Equilibrium  for  Any  Part  of  Cord  Carry- 
ing Distributed  Vertical  Load. —  Let  it  be  required  to  determine 
the  form  of  the  curve  assumed  by  a  cord  suspended  from  two  fixed 
points  and  sustaining  a  vertical  load  distributed  along  the  cord  in 
any  manner. 

Let  the  coordinates  of  any  point  of  the  curve  (  Fig.  82)  be  x  and  y, 
the  axis  of  x  being  horizontal  and  that  of  y  vertical,  and  the  origin 
being  any  point  in  the  plane  of  the  curve.  Let  cf>  denote  the  angle 
between  the  tangent  to  the  curve  at  any  point  and  the  axis  of  x,  and 
T  the  tension  in  the  cord  at 
that  point.  Also  let  w  be 
the  load  sustained  by  the 
cord  per  unit  length  (made 
up  of  its  own  weight  and  any 
other  applied  load).  In  gen- 
eral w  varies  continuously 
along  the  cord. 

Let  P  and  Q  be  any  two 
points  of  the  cord,  and  let 
the  values  of  x%  y,  T  and  <f> 
at  P  be  denoted  by  xlyyi} 
7\ ,  (/>t  ;  while  at  Q  the  val- 
ues are  x2 ,  y, ,  T2 ,  <f}r  Let  W  represent  the  total  load  applied  to 
the  portion  PQ.  Applying  the  conditions  of  equilibrium  to  PQ,  and 
resolving  forces  horizontally, 

T2  cos  (f>2  —  7",  cos  (f>l  =  o;     .         .         .     (1) 
resolving  vertically, 

T2  sin  <f>,  —  Tx  sin  <^>1  —  ^f=o.  .  .      (2) 

A  moment  equation  might  also  be  written,  but  is  not  needed  in 
the  present  discussion. 

Equation  (1)  expresses  the  fact  that  the  horizontal  component  of 


''*V'  4h 


O 


Fig.  82. 


Il8  THEORETICAL    MECHANICS. 

the  tension  has  the  same  value  at  all  points  of  the  cord.  Let  this 
value  be  represented  by  H,  so  that 

Tx  cos  (/>j  =  T%  cos  (j>2  =  T  cos  </>  =  H  =  constant.    .      (3) 

Substituting  in  equation  (2)  the  values  Tx  =  Hj cos  <f>lt 
T2  =  HI  cos  </>2 ,  that  equation  becomes 

tan  (f>t  —  tan  </>,  =  WjH.         .         .         .      (4) 

Equations  (3)  and  (4)  may  be  used  instead  of  (1)  and  (2). 

If  the  values  of  T  and  </>  are  known  at  any  one  point  of  the  cord, 
H  becomes  known.  If  also  W  is  known  for  every  portion  of  the 
cord,  equation  (4)  serves  to  compute  the  slope  of  the  curve  at  any 
point.  A  differential  equation  may,  however,  be  deduced  which  is 
more  useful. 

140.  Differential  Equation  of  Curve  of  Loaded  Cord. — Let  s 

denote  the  length  of  the  curve  measured  from  some  fixed  point  up 
to  the  point  (x,  y),  and  let  sx  and  s%  be  the  values  of  s  at  P  and  Q. 
From  equation  (4)  we  have 

(tan  &  -  tan  fa)/  (s.z  -  st)  =  W/H  (s.z  -  *,)• 

Let  the  limiting  value  of  each  member  of  this  equation  be  found  as 
P  and  Q  approach  coincidence.      Evidently, 

limit  [(tan  (j>2  —  tan  </>x)  /  (st  —  sx)]  =  ^(tan  <f>)/ds; 

limit  [W/(s2  —  sx)]  =  w. 

Hence  the  equation  becomes 

d(tan<f>)/ds  =  wjH.        .        .        .     (5) 

To  apply  this  equation  to  any  particular  case,  the  distribution  of 
loading  must  be  known,  so  that  w  can  be  expressed  in  terms  of  the 
coordinates  of  the  curve.     Two  important  cases  will  be  considered. 

141.  The  Common  Catenary.— The  curve  assumed  by  a  cord 
whose  weight  per  unit  length  is  uniform,  is  called  the  common  cat- 
enary. 

In  this  case  w  is  constant ;  let  Hj  w  =  c,  so  that  the  constant  c 
denotes  the  length  of  cord  whose  weight  is  equal  to  the  horizontal 
tension  H.     Writing/  for  tan  $,  equation  (5)  becomes 

dp  jds  =  ijc.         .         .  .  .      (6) 


EQUILIBRIUM    OF    FLEXIBLE    CORDS. 


119 


Since  dp  Ids  =  {dp /  dx){dx /  ds)  ==   (dpjdx)  cos  <f>,  and  cos  </>  = 
i/l/  1  -j-  /2,  equation  (6)  may  be  written 


Integrating, 


dp/Vi  +  p2  =  dx/c. 


(7) 


log(/  +  l/i+/*)=  */'+*". 


To  determine  the  constant  ^T,  suppose  the  tangent  of  the  curve  at 
some  point  to  be  horizontal,  and  let  the  origin  of  coordinates  lie  in 
a  vertical  line  through  that  point,  so  that  when  x  equals  zero,  p  equals 
zero.     Then  K  equals  zero,  and  the  equation  may  be  written 


*x\c 


Solving  for  p, 
Integrating, 


p  =  dy/dx  ==  y2{exic  —  e-*'c\ 
y  =  y2c(ex!c  -f  e-*'c)  +  K'. 


•  (8) 

•  (9) 
.   (10) 


If  the  origin  of  coordinates  is  taken  at  the  lowest  point  of  the 
curve,  so  that  y  =  o  when  x 
=  o,  K'  is  equal  to  — c. 
If  the  origin  is  at  a  distance 
c  below  the  lowest  point  of 
the  curve,  K'  =  o.  Adopt- 
ing the  latter  supposition, 
the  equation  becomes 

;=  %c(e*l*  +  e~xlc\  (11) 

In  the  solution  of  prob- 
lems relating  to  the  catenary, 
use  is  made  of  certain  rela- 
tions between  the  following 
quantities  :  the  length  of  the 
cord ;    the    tension    at    any 

point ;  the  tension  at  the  lowest  point ;  the  inclination  of  the  tangent 
at  any  point ;  the  coordinates  of  any  point.  The  following  equa- 
tions may  be  deduced : 

(a)  From  equation  (6),  by  integration, 


Fig.  83. 


s  =  cp  =  c  tan  <f> 


(12) 


the  constant  of  integration  being  zero  if  s  is  reckoned  from  the  point 
at  which  (f>  =  o,  that  is,  from  the  lowest  point  of  the  curve. 


120  THEORETICAL    MECHANICS. 

(/?)  Eliminating  x  between  equations  (9)  and  (11),  there  results 


y  =  cv  1  -\- p2  ==  ^sec  <f>.      .         .   •      .   (13) 
(V)  Eliminating/  between  (12)  and  (13), 

r-s'=c2 (14) 

The  use  of  these  equations  will  now  be  illustrated. 

Examples. 

1 .  Let  a  cord  be  suspended  from  two  points  at  a  given  distance 
apart  in  the  same  horizontal  line,  the  load  per  unit  length  (w)  being 
known,  and  the  tension  at  the  lowest  point  (//)  being  assigned. 
Required  (1)  the  length  of  the  curve,  (2)  the  position  of  the  lowest 
point,  (3)  the  slope  at  points  of  suspension,  (4)  the  tension  at  points 
of  suspension. 

Since  w  and  H  are  given,  c  is  known,  its  value  being  77/ w. 
Also,  the  value  of  x  at  a  point  of  suspension  is  known  (being  half  the 
distance  between  the  two  given  points);  hence  from  (11)  the  value 
ofjj/at  a  point  of  suspension  can  be  computed.  Subtracting  cfrom 
this  value  of  y  gives  the  depth  of  the  lowest  point  below  the  points  of 
suspension  ;  which  answers  question  (2). 

The  length  of  half  the  curve  may  be  found  by  substituting  in 
equation  (14)  the  value  of  c  and  the  value  ofy  at  the  point  of  sus- 
pension.    This  answers  question  (1). 

From  (13)  may  be  found  the  value  of  </>  at  the  point  of  suspension, 
which  answers  question  (3). 

From  equation  (3), 

T  —  //sec  <f>, 

from  which  may  be  determined  the  value  of  T  at  the  points  of  sus- 
pension, thus  answering  question  (4). 

In  solving  a  numerical  case  of  this  example,  the  value  of  the 
exponential  terms  in  equation  (11)  must  be  found  by  logarithms. 
The  value  of  e,  the  base  of  the  system  of  hyperbolic  logarithms,  is 
2. 7 1 828 1 8  and  its  common  logarithm  is  0.4342945. 

2.  Let  the  points  of  suspension  be  100  ft.  apart  in  the  same  hori- 
zontal plane,  let  the  load  per  foot  be  50  lbs.,  and  let  it  be  required 
that  the  tension  at  the  lowest  point  shall  be  1000  lbs. 

From  the  given  data, 

c  =  Hjw  =  1000/50  =  20  ft. 

The  coordinates  of  one  of  the  points  of  suspension  are 

x  =  50  ft.,     y  =  10  O2-5  -J-  *-25)  =122.6  ft. 

The  lowest  point  of  the  curve  is  therefore   102.6  ft.  lower  than  the 
points  of  suspension. 


EQUILIBRIUM    OF    FLEXIBLE    CORDS.  121 

For  the  half  length  of  the  curve, 

s2  =  y*  —  c2  =  (122.6)2  —  (20)2; 
s  =  1 2 1 .0  ft. 
For  the  value  of  (f>  at  the  point  of  suspension  we  have 

cos  (f>  =  c/y  =  20/122.6  =  o.  163 1  ;    </>  =  8o°  37'. 
The  value  of  T  at  the  end  is 

T==  H  sec  <\>  =  6, 130  lbs. 

3.  A  cord  of  known  length  is  suspended  from  two  given  points 
in  the  same  horizontal  plane  ;  determine  the  position  of  the  lowest 
point  of  the  curve,  and  the  tension  at  any  point. 

In  this  case  the  value  of  x  at  the  end  of  the  cord  is  known,  but 
the  corresponding  value  of  y  is  unknown,  as  is  also  the  value  of  c. 
To  solve  the  problem,  y  and  c  must  be  determined  so  as  to  satisfy 
equations  (n)  and  (14).  This  is  best  accomplished  by  trial  when 
numerical  data  are  given. 

142.  Load  Distributed  Uniformly  Along  the  Horizontal. —  Let 

w'  denote  the  load  per  unit  horizontal  distance.  Since  the  load  on 
a  length  ds  of  the  curve  falls  upon  a  horizontal  distance  dx, 

wds  =  w'dx, 

or  w'  =5  w{ds/dx)  =  w  sec  $.  .         .     (15) 

Consider  now  the  case  in  which  w'  is  constant. 

Writing/ for  tan  <j>,  and  substituting  w '  cos  (j>  for  w,  equation  (5) 
(Art.  140)  may  be  written 

dp jds  =  (w'  cos  <\>)IH  =  {cos  </>)/r',      .         .     (16) 

in  which  c'  is  written  for  11/ w',  and  is  a  constant  denoting  the  hori- 
zontal distance  upon  which  the  load  is  equal  to  the  horizontal  tension 
H.  Again,  since  dp/ds  =  (dp/dx)(dx/ds)  =  (dp/dx)  cos  (/>,  the 
equation  becomes 

dp/dx  =  d2yldx*  =  i/c'.      .         .         •     (17) 

Integrating,  taking  the  axis  of  y  through  the  point  where  the  curve 
is  horizontal,  so  that  when  x  =  o,  dy/dx  =  o,  there  results 

p  =  dy/dx  =  x/cf.  .         .  .     (18) 

Integrating  again,  taking  the  origin  on  the  curve,  so  that  y  =  o 
when  x  =  o, 

y=x2/2c' (19) 

This  represents  a  parabola  with  vertex  at  the  origin  of  coordinates  and 
axis  vertical. 


ds  =  c'V \  +  p2-  dp, 


122  THEORETICAL    MECHANICS. 

The  length  of  the  curve,  measured  from  the  origin  to  any  point, 
may  be  determined  as  follows  : 

Since  cos  <j>  a=  l/vj  -f-  p2,  equation  (16)  may  be  written 

ds  =  c'V 
the  integral  equation  of  which  is 

s  =  y*cXpVi  -\-p2  +  log  (p  +  Vi  +/»)],     .     (20) 

the  constant  of  integration  being  zero  if  s ..ass  o  when  p  =  o.  If  it  is 
desired  to  express  s  in  terms  of  x,  p  may  be  eliminated  by  means 
of  equation  (18). 

Examples. 

1.  A  cord  carrying  a  known  load  with  uniform  horizontal  distri- 
bution is  suspended  from  two  points  on  the  same  level  at  a  given 
distance  apart.  The  lowest  point  of  the  cord  is  at  a  known  distance 
below  the  points  of  suspension.  To  determine  (a)  any  number  of 
points  of  the  curve  ;  (b)  the  tension  at  any  point ;  (c)  the  length  of 
the  cord. 

(a)  Since  the  values  of  x  and  y  at  the  points  of  suspension  are 
known,  c  can  be  determined  from  equation  (19);  after  which  the 
same  equation  serves  to  determine  the  coordinates  of  any  number  of 
points  of  the  curve. 

(b)  The  value  of  w'  being  known,  //"can  be  computed  from  the 
relation  H  =  c'w'.  The  slope  at  any  point  may  be  found  from  the 
equation  tan  £  =/  =  dy\dx  =  xlc\ 

and  the  tension  at  any  point  by  the  relation 

T  =  H  sec  </>  a=  H  V\  +f. 

(V)  The  length  of  the  cord  from  the  lowest  point  up  to  any  given 
point  may  be  found  from  equation  (20),  using  the  above  value  of  p. 

(2)  A  cord  carrying  40  lbs.  per  horizontal  foot  is  suspended  at 
two  points  on  the  same  level  40  ft.  apart,  so  that  the  lowest  point  of 
the  cord  is  10  ft.  below  the  point  of  suspension.  Required  the  ten- 
sion at  the  lowest  point ;  the  tension  at  a  point  of  suspension  ;  and 
the  length  of  the  cord. 

143.  Approximate    Solution  of    Problem   of   Loaded   Cord. — 

If  a  cord  is  suspended  from  two  points  on  the  same  level  whose  dis- 
tance apart  differs  little  from  the  length  of  the  cord,  the  relation 
between  the  load,   the  sag,*  the  distance  between  points  of  sus- 

*  By  the  sag  is  meant  the  vertical  distance  of  the  lowest  point  of  the  cord 
below  the  level  of  the  points  of  support. 


EQUILIBRIUM    OF    FLEXIBLE    CORDS.  1 23 

pension  and  the  length  may  be  determined  approximately  as 
follows : 

If  the  weight  per  unit  length  is  uniform,  it  may  be  assumed,  with 
small  error,  that  the  load  is  uniformly  distributed  along  the  hori- 
zontal, and  that  w'  =  w.  On  these  assumptions  the  curve  is  a  para- 
bola and  equations  (19)  and  (20)  of  Art.  142  are  applicable.  An 
approximate  value  of  s  may  be  obtained  in  simple  form. 

If  the  value  of  s  given  by  equation  (20)  be  developed  in  powers 
Oi  /,  the  first  two  terms  of  the  result  are 

This  result,  expressed  in  terms  of  x,  is 

s  =  c'{xlc'  +  x3[6cn)  =  *  (1  -f  xV6cn). 

Examples. 

1.  A  wire  weighing  o.  1  lb.  per  "yard  of  length  is  suspended  be- 
tween two  points  100  ft.  apart  so  that  the  sag  at  the  middle  is  2  ft. 
Determine  the  length  and  the  greatest  tension. 

The  equation  of  the  curve  being  y  =  x2/2Cf,  the  value  of  c', 
found  by  substituting  in  this  equation  the  coordinates  of  any  point  of 
the  curve,  is 

cr  as  x2l2y  =  2500/4  =  625  ft. 

The  tension  at  the  lowest  point  is 

H  =  c'w'  =  625/30  3=  20.8  lbs. 
The  value  of  p  at  the  support  is 

p  =  x/c'  =  50/625  =  0.08, 
and  the  value  of  the  tension  at  the  point  of  suspension  is  therefore 

T  —  H  sec  </>  =  20.8V  1  -f-  0.0064  =  2°-9  l°s- 
The  half  length  of  the  curve  is 

s  =  c'(p  +  ps/6)  =  625  (0.08  -j-  0.000512/6)  =  50.053  ft. 

2.  Solve  Ex.  1,  assuming  the  sag  equal  to  (a)  1  ft.  and  (b)  3  ft. 

144.  Suspension  Bridge. — The  cables  of  a  suspension  bridge 
usually  hang  in  parabolic  curves.  If  the  total  load  carried  by  such 
a  cable  has  a  uniform  horizontal  distribution,  there  is  no  tendency  of 
the  cables  to  depart  from  the  parabolic  form.  If,  however,  this  dis- 
tribution of  loading  is  departed  from,  there  is  a  tendency  to  change 
the  form  of  the  curve  assumed  by  the  cable.  To  counteract  this 
tendency  to  distortion  is  the  office  of  the  "stiffening  truss." 

For  a  complete  discussion  of  the  theory  of  suspension  bridges, 
works  on  the  theory  of  bridges  must  be  consulted. 


CHAPTER    IX. 

CENTROIDS. 

§  i.    Centroid  of  a  System  of  Parallel  Forces. 

145.  Forces  with  Fixed  Points  of  Application. —  In  the  discus- 
sion of  systems  of  forces  applied  to  rigid  bodies,  it  has  been  assumed 
that  a  force  may  be  applied  at  any  point  in  its  line  of  action,  since  its 
effect  upon  the  motion  (or  tendency  to  motion)  of  the  body  is  the 
same  for  all  such  points  of  application.  In  certain  cases,  however, 
forces  are  applied  at  definite  points  which  remain  fixed  in  the  body, 
whatever  displacement  it  may  undergo.  Thus,  the  force  of  gravity 
acting  upon  a  body  is  the  resultant  of  forces  applied  to  every  particle 
of  the  body  and  these  points  of  application  remain  fixed  in  the  body, 
however  it  may  be  turned  from  its  original  position. 

In  the  following  discussion  of  systems  of  forces  with  points  of 
application  fixed  in  the  body,  it  will  be  assumed  that  the  forces  are 
parallel,  and  that  they  remain  constant  in  magnitude  and  direction, 
however  the  body  may  be  displaced. 

It  obviously  amounts  to  the  same  thing,  so  far  as  the  relations  of 
the  forces  are  concerned,  whether  their  direction  remains  fixed  while 
the  body  turns,  or  whether  their  direction  changes  while  the  body 
remains  fixed. 

146.  Resultant  of  Two  Parallel  Forces  with  Fixed  Points  of 
Application. —  Let  two  parallel  forces  be  applied  at  fixed  points,  A 
and  B.  Their  resultant  is  equal  to  their  algebraic  sum,  and  its  line 
of  action  divides  the  line  A B  into  segments  inversely  proportional  to 
the  given  forces  (Art.  86).  Let  C  (Fig.  84)  be  the  point  in  which 
the  line  of  action  of  the  resultant  intersects  AB.  If  the  two  given 
forces  remain  parallel  while  their  direction  changes  in  any  manner, 
their  points  of  application  still  being  A  and  B,  the  line  of  action  of 
the  resultant  will  continue  to  pass  through  the  same  point  C ;  hence 
C  may  be  taken  as  the  fixed  point  of  application  of  the  resultant. 

147.  Resultant  of  Any  Number  of  Parallel  Forces  with  Fixed 
Points  of  Application.—  If,  in  addition  to  the  two  parallel  forces 
applied  at  A  and  73,  there  is  a  third  force  parallel  to  them  applied  at 


CENTROIDS.  125 

a  point  D,  the  resultant  of  the  three  may  be  found  by  combining  the 
third  with  the  resultant  of  the  first  two.      Since  the  resultant  of  the 
first  two  has  a  fixed  point  of  applica- 
tion for  all  directions  in  which   the 
forces  may  act,  and  since  a  point  may 
also  be  found  which  may  be  regarded 

as  the  fixed  point  of  application  of       1  y  /    v  *  -  „    2 

the  resultant  of   this  force  and    the 
third  of  the  given  forces,  it  follows 


R 


that  the  line  of  action  of  the  resultant 

of  the  three  given  forces  will  always 

pass  through  a  certain  point  (as  E, 

Fig.  84),  which  may  be  taken  as  its 

fixed  point  of  application.     Since  this  process  may  be  continued  to 

include  any  number  of  forces,  it  is  seen  that  the  resultant  of  any 

number  of   parallel  forces  with  fixed  points  of  application  acts  in 

a  line  which  always  passes  through  a  certain  fixed  point,  whatever 

the  direction  of  the  forces. 

148.  Centroid. —  The  point  of  application  of  the  resultant  of  any 
system  of  parallel  forces  with  fixed  points  of  application  is  called  the 
centroid  of  the  system.  Methods  of  determining  the  centroid  will 
now  be  considered. 

149.  Determination  of  Centroid  of  Coplanar  Forces. —  Let  7^, 

P„ ,     .  represent  the  given  forces,  and  let  the  coordinates  of 

their  points  of  application  with  reference  to  a  pair  of  rectangular  axes 
be  (xx ,  yd*  (x-i  *y*)>  -  -  -  -  Let  R  denote  the  resultant  and  x,  y  the 
coordinates  of  its  point  of  application,  that  is,  the  coordinates  of  the 
centroid  of  the  system.  Assume  the  forces  to  act  first  parallel  to  the 
axis  of  y  and  then  parallel  to  the  axis  of  x.  The  magnitude  of  the 
resultant  is  the  same  in  both  cases,  and  in  each  case  it  acts  in  a  line 
which  contains  the  centroid.  Since  the  moment  of  the  resultant  of 
any  system  of  forces  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  several  forces,  we  may  write,  for  the  case  in  which  the  forces  act 
parallel  to  the  axis  of  y}  the  equation 

Rx  =  Pxxx  +  Ptx%  +   \      .      .      ; 

and  for  the  case  in  which  they  act  parallel  to  the  ^r-axis, 

*?-/Wr+"J&.+     •      •     •     • 


126  THEORETICAL    MECHANICS. 

From  these,  since  R  is  the  algebraic  sum  of  the  given  forces,  the 
coordinates  of  the  centroid  are  found  to  be 

X  =  (PXXX  +  P2X2+     .     .     .     )/(/>    +/*+:.,,)==  ^PX^P\ 

y  =  tp^  +  i\y,+  .  .  .  )/(/;  + /y+  .  .  .  )=2/>/sp. 

Here  the  sign  2  denotes  the  summation  of  a  series  of  terms  of  sim- 
ilar form. 

150.  Non-Coplanar  Parallel  Forces. —  It  will  be  well  to  extend 
the  discussion  to  cases  in  which  the  forces,  though  parallel,  are  not 
restricted  to  a  plane  and  the  points  of  application  have  any  positions 
in  space. 

The  reasoning  of  Arts.  146  and  147,  showing  that  every  system 
of  parallel  forces  with  fixed  points  of  application  has  a  centroid,  ap- 
plies to  this  general  case.  To  determine  the  position  of  the  centroid, 
the  following  method  may  be  employed. 

(a)  Coplanar  points  of  application. — If  the  points  of  application 
are  coplanar,  while  the  direction  of  the  forces  is  unrestricted,  the 
centroid  may  be  determined  by  assuming  the  forces  to  act  in  the 
plane  of  the  points  of  application.  If  the  axes  of  x  and  y  are  chosen 
in  this  plane,  the  coordinates  of  the  centroid  are  given  by  the  formulas 
deduced  for  coplanar  forces. 

(b)  Non-coplanar  points  of  application. —  Let  slt  z2,  etc.,  denote 
the  ordinates  of  the  points  of  application  of  the  several  forces  meas- 
ured from  any  assumed  reference  plane.  Any  two  of  the  given 
points  of  application  lie  in  a  plane  perpendicular  to  the  reference 
plane ;  hence  the  formula  deduced  for  the  case  of  coplanar  points  of 
application  applies  to  any  two  forces.  Or,  if  z'  is  the  ordinate  of  the 
centroid  of  Px  and  P2 , 

Similar  reasoning  applies  to  the  two  forces  (Pt  -f-  P2)  and  P3 ;  if  z" 
denotes  the  ordinate  of  the  centroid  of  these  forces, 

*'  =  [(/>  +  />,)  *'  +  /»,.*,]/[(/>,  +  /»)  +  PZ 

=  (P.ir,  +  P,s,+  Pt*,)KP,  +  P%  +  P3). 

By  extending  this  reasoning,  any  number  of  forces  may  be  included, 
with  the  result  that  the  ordinate  of  the  centroid  is 

j?  =  (/y*,  +  ^2+  .  .  .  )/(/>,  +  p2  +  .  .  .  )  =  zpz/zp. 


CENTROIDS.  127 

Since  the  plane  of  reference  may  be  any  plane  whatever,  an  indefi- 
nite number  of  equations  of  the  above  form  may  be  written.  Three 
equations,  however,  suffice  for  the  determination  of  the  centroid.  If 
three  rectangular  coordinate  planes  are  selected,  and  the  points  of 
application  of  the  forces  are  (xlyyA,  zx),  (x-i,y<i,  z2)>  etc->  tne  coor- 
dinates of  the  centroid  are  given  by  the  three  equations 

x  =  2,Pxp.P;    y  =  ZPy/ZP;     Z  =  ZPs/ZP 


§  2.    Centroids  of  Masses,    Volumes,  Areas  and  Lines. — General 

Method. 

151.  Center  of  Mass  Defined. —  If  parallel  forces  be  conceived 
to  act  in  the  same  direction  upon  all  portions  of  a  body,  such  that 
the  resultant  forces  acting  upon  any  two  portions,  however  small, 
are  proportional  to  their  masses,  the  centroid  of  this  system  of  forces 
is  called  the  center  of  mass  of  the  body. 

The  forces  of  gravity  acting  upon  all  portions  of  a  body  form  a 
system  which  is  practically  such  as  described  in  this  definition.  The 
centroid  of  the  forces  of  gravity  is  called  the  center  of  gravity  of  the 
body.  The  terms  center  of  mass  and  center  of  gravity  are  often 
used  interchangeably,  although  it  is  to  be  remembered  that  the  forces 
of  gravity  do  not  exactly  satisfy  the  conditions  assumed  in  the  defini- 
tion of  center  of  mass. 

The  determination  of  the  mass-center  of  a  body  or  of  a  system  of 
bodies  is  a  special  case  of  the  determination  of  the  centroid  of  a 
system  of  parallel  forces.  The  word  centroid  is,  in  fact,  often  used 
to  designate  the  center  of  mass. 

152.  General  Method  of   Determining  Center  of   Mass. —  Let 

the  given  body  be  divided  into  any  number  of  parts  whose  masses 
areWp  mai  .  .  .  ,  and  let  the  coordinates  of  their  respective  centers 
of  mass  be  (xlf  ylt  zx),  (x.it  yt  ,zi),  ....  Assuming  forces  such 
as  described  in  the  definition  of  mass-center  (Art.  151)  to  act  upon  all 
portions  of  the  body,  the  problem  is  to  find  the  point  of  application 
of  the  resultant  of  these  forces.  Now,  from  the  definition  of  mass- 
center  it  follows  that  the  forces  acting  upon  all  parts  of  tnx  may  be 
replaced  by  their  resultant,  taken  as  applied  at  the  mass-center  of 
tnx  ;  and  similarly  with  each  of  the  other  portions.  Hence,'if  x,  y}  z 
are  the  coordinates  of  the  mass-center  of  the  whole  body,  there  may 


128  THEORETICAL    MECHANICS. 

be  written  three  equations  similar  to  those  which  give  the  cooordi- 
nates  of  the  centroid  of  any  system  of  parallel  forces  (Art.  150); 
the  force-magnitudes  being  replaced  by  the  masses  mlt  m.n  .  .  .  . 
The  equations  are,  therefore, 

x  =  ^mx/^m  ;    y  =  'Zmy/'Zm  ;     z  =  ^mz  /2w. 

To  apply  these  formulas,  it  is  necessary  to  know  the  center  of 
mass  of  each  of  the  separate  masses.  If  the  given  body  can  be  sub- 
divided into  finite  parts  whose  masses  and  mass-centers  are  known, 
the  application  of  the  formulas  is  simple.  If  such  subdivision  is 
impossible,  resort  must  be  had  to  the  integral  calculus,  if  an  exact 
determination  is  required. 

The  above  formulas  are  directly  applicable  if  it  is  desired  to  de- 
termine the  center  of  mass  of  any  number  of  bodies  taken  together, 
their  several  masses  and  mass -centers  being  known. 

Examples. 

1.  Find  the  center  of  mass  of  a  system  of  four  bodies  whose 
masses  are  10  lbs.,  50  lbs.,  12  lbs.,  40  lbs  ;  their  several  mass-centers 
being  at  the  points  whose  rectangular  coordinates  are  (10,  4,  • — 3), 
(12,  —3,  2),  (4,  —5,  —6),  (—8,  3,  6). 

2.  Show  that  the  mass-center  of  three  particles  of  equal  mass 
placed  at  the  vertices  of  a  triangle  lies  at  the  intersection  of  the 
medians. 

3.  Show  that  the  mass-center  of  four  particles  of  equal  mass 
placed  at  the  vertices  of  a  triangular  pyramid  is  at  a  distance  from 
the  base  equal  to  one-fourth  the  altitude. 

153.  Density. —  A  body  is  said  to  be  homogeneous  if  the  masses 
of  any  two  portions,  however  small,  are  proportional  to  their  volumes. 

The  density  of  a  homogeneous  body  is  a  quantity  proportional 
directly  to  the  mass  and  inversely  to  the  volume.  If  the  unit  density 
is  that  of  a  body  of  which  unit  volume  contains  unit  mass,  the  den- 
sity of  any  homogeneous  body  is  equal  to  the  mass  of  unit  volume 
of  the  body. 

If  p  denotes  the  density,  V  the  volume  and  M  the  mass, 

p  =  M/V,     or    M=  pV, 

for  a  homogeneous  body. 

Dimensions  of  unit  density. — The  unit  density  above  described 
is  a  derived  unit  (Art.  13),  depending  upon  the  units  of  mass  and 


CENTROIDS.  129 

length.  This  dependence  is  expressed  by  the  dimensional  equation 
(Art.  15) 

(unit  density)  ===  (unit  mass)/(unit  volume)  =  M/L3. 

If  the  pound  is  taken  as  the  unit  mass  and  the  foot  as  the  unit  length, 
the  unit  density  is  that  of  a  body  of  which  each  cubic  foot  of  volume 
contains  one  pound  mass.  The  density  of  any  body  is  then  ex- 
pressed as  a  certain  number  of  pounds  per  cubic  foot.  Other  units 
of  mass  and  length  give  other  units  of  density,  —  for  example  a  gram 
per  cubic  centimeter,  or  a  kilogram  per  cubic  meter.  The  methods 
of  determining  center  of  mass  are  independent  of  the  particular  unit 
of  density. 

Average  density. — If  a  body  is  not  homogeneous,  its  average 
density  may  be  defined  as  the  density  of  a  homogeneous  body  whose 
total  mass  and  total  volume  are  equal  to  those  of  the  given  body. 
The  average  density  is  thus  equal  to  the  whole  mass  divided  by  the 
whole  volume. 

Actual  density  at  a  point. — In  case  of  a  heterogeneous  body, 
the  average  densities  of  different  portions  are  unequal.  The  concep- 
tion of  actual  density  at  a  point  may  be  reached  as  follows  : 

Let  AM  be  the  mass  and  A  V  the  volume  of  any  portion  con- 
taining the  given  point ;  its  average  density  p'  is  then 

p;  =  AM/AV. 

If  A  V  become  smaller,  approaching  zero  as  a  limit,  but  always  con- 
taining the  given  point,  AM  also  generally  approaches  zero,  and  p' 
approaches  a  definite  limit  p.  This  limiting  value  of  p'  is  the  den- 
sity at  the  point  considered.  That  is,  the  required  density  is  given 
by  the  equation  p  ^  dMjdVt 

154.  Mass-Center  of  Homogeneous  Body  or  System  of 
Bodies. —  For  a  homogeneous  body,  or  for  a  system  of  such  bodies 
of  equal  densities,  volumes  may  be  substituted  for  masses  in  the  for- 
mulas for  the  coordinates  of  the  mass-center.  For  if  p  is  the  den- 
sity of  each  body,  and  vx ,  vt\  .  .  .  are  the  volumes  correspond- 
ing to  masses  mx  , ffht     •     •     •      ,  we  have 

ml  =  pvx ,     m2  =  pv2i    .     .     .     ; 

and  substituting  these  values  in  the  formulas  for  x,  y}  and  2,  the  factor 
p  cancels,  leaving 

x  =  1vx/1v ;    y  =  ^vy/^v ;     2  =  lLvz[S*v. 


I30  THEORETICAL    MECHANICS. 

155.  Centroids  of  Volumes,  Surfaces  and  Lines. —  Volumes. — 
If  parallel  forces  be  conceived  to  be  applied  to  all  portions  of  a  body, 
the  resultant  forces  acting  upon  any  two  portions,  however  small, 
being  proportional  to  their  volumes,  the  centroid  of  this  system  of 
forces  is  called  the  center  of  volume  (or  volume- centroid}  of  the  body. 

The  centroid  of  a  given  volume  evidently  coincides  with  the 
mass-center  of  a  homogeneous  body  occupying  the  volume,  and  may 
be  found  by  the  formulas  already  given. 

Surfaces. — The  centroid  (or  center  of  area)  of  any  surface  is  the 
centroid  of  a  system  of  parallel  forces  conceived  to  be  applied  to  all 
portions  of  the  surface,  the  resultant  forces  acting  upon  any  two  por- 
tions, however  small,  being  proportional  to  their  areas.  This  defini- 
tion applies  to  curved  surfaces  as  well  as  to  plane  areas.  In  case  of 
a  plane  surface  the  centroid  lies  in  the  plane  ;  but  the  centroid  of  a 
curved  surface  does  not  in  general  lie  in  the  surface. 

The  formulas  for  the  coordinates  of  the  centroid  of  a  homogeneous 
body  or  of  a  volume  may  be  applied  in  finding  the  centroid  of  an 
area,  if  vlt  tf»,  .  .  .  represent  partial  areas  and  (xx,  yx,  2X)> 
(•*2  >  y*  1  #*)     .      .      .      the  coordinates  of  their  centroids. 

Lines. — The  centroid  (or  center  of  length)  of  any  line  is  the  cen- 
troid of  a  system  of  parallel  forces  conceived  to  be  applied  to  every 
portion  of  the  line,  the  resultant  forces  acting  upon  any  two  portions, 
however  small,  being  proportional  to  their  lengths. 

If  the  line  is  divided  into  parts  whose  lengths  and  centroids  are 
known,  the  centroid  of  the  whole  line  is  found  by  formulas  identical  in 
form  with  those  used  for  finding  the  coordinates  of  the  centroid  of  a 
volume  or  area  ;  but^,  v2,     .      .      .     must  represent  lengths. 

156.  Moment  with  Respect  to  a  Plane. —  The  moment  of  a  mass 
with  respect  to  a- plane  is  defined  as  the  product  of  the  mass  into 
the  distance  of  its  mass-center  from  the  plane. 

The  moment  of  a  volume  with  respect  to  a  plane  is  the  product 
of  the  volume  into  the  distance  of  its  centroid  from  the  plane. 

The  moment  of  a  surface  (or  of  a  line)  with  respect  to  a  plane 
is  the  product  of  its  area  (or  length)  into  the  distance  of  its  centroid 
from  the  plane. 

The  formula  for  the  jr-coordinate  of  the  centroid  of  any  mass 
(Art.  152)  maybe  written 

{mx  -f  m2  -j-  .     .     .     )x  =  mx  xx  +  m2  x%  +     ,     ,     .     . 


CENTROIDS.  131 

Since  the  plane  from  which  x  is  measured  may  be  any  plane,  the 
equation  expresses  the  proposition  that 

The  moment  of  any  mass  with  respect  to  a  plane  is  equal  to  the 
sum  of  the  moments  of  any  parts  into  which  it  may  be  divided. 

The  same  proposition  obviously  holds  for  a  volume,  a  surface  or 
a  line. 

As  a  special  case,  if  the  given  plane  contains  the  centroid  of  the 
body,  surface  or  line,  the  sum  of  the  moments  of  the  parts  is  zero. 

I57«  Symmetry. —  If  a  volume,  a  surface  or  a  line  has  a  plane 
of  symmetry,  the  centroid  lies  in  this  plane.  For  the  volume,  sur- 
face or  line  can  be  divided  into  pairs  of  equal  elements  such  that  the 
centroid  of  each  pair  lies  in  the  plane  of  symmetry. 

Similarly,  if  there  is  an  axis  of  symmetry  it  contains  the  centroid  ; 
and  if  there  is  a  center  of  symmetry  it  coincides  with  the  centroid. 

A  similar  proposition  is  true  for  a  mass  if  homogeneous. 

From  the  principles  of  symmetry  the  centroids  of  many  geometri- 
cal figures  may  be  located,  either  completely  or  partly,  by  inspection. 
Thus : 

The  centroid  of  a  straight  line  is  at  its  middle  point. 

The  centroid  of  a  circular  arc  lies  upon  the  line  bisecting  the 
angle  subtended  at  the  center. 

The  centroid  of  a  rectangular  area  lies  upon  the  line  bisecting  two 
opposite  sides,  and  is  therefore  at  the  intersection  of  the  two  lines  so 
drawn. 

The  centroid  of  the  area  of  an  ellipse  is  at  its  center  of  figure. 

The  centroid  of  an  ellipsoid  is  at  its  center. 

158.  Centroids  of  Plane  Figures  and  of  Volumes. — Parallelo- 
gram.—  Let  A  BCD  (Fig.  85)  be  a  par- 
allelogram.     Since  the  relation  of  the  A. Lj B 

area  to  AB  and  its  relation  to  the  oppo- 
site side  CD  are  exactly  similar,  the 
centroid  is  equally  distant  from  these 
two  lines.  For  a  like  reason  it  is  equal- 
ly distant  from  AD  and  BC.  It  is  there-  j}  ~M 
fore  at  the  intersection  of  the  bisectors  pIG  8s 
LM,  UK.     This  point  of  intersection 

is,  in  fact,  the  center  of  symmetry  of  the  parallelogram.  It  is  also 
the  point  of  intersection  of  the  diagonals. 


132  THEORETICAL    MECHANICS. 

Triangle. —  In  the  triangle  ABC  (Fig.  86)  draw  AX  bisecting 
the  side  BC,  and  draw  lines  such  as  B'  C,  parallel  to  BC  and  lim- 
ited by  AB  and  A  C.  These  lines  are  all  bisected  by  AX.  Draw 
B  'b  and  C'c parallel  to  AX,  thus  constructing  a  parallelogram  B '  C'cb. 
The  centroid  of  every  such  paralelogram  lies  on  A  X,  hence  the  cen- 
troid  of  the  figure  made  up  of  all  of  them  lies  on  AX.  If  the  num- 
ber of  the  parallelograms  be  increased  indefinitely,  the  width  of  each 

approaching  zero,  this  figure  approaches 
coincidence  with  the  triangle  ;  hence  the 
centroid  of  the  triangle  lies  upon  AX. 
Similar  reasoning  shows  that  it  lies  on  the 
line  drawn  from  C  bisecting  AB,  and  on 
the  line  drawn  from  B  to  the  middle  point 
of  A  C.  Hence  it  is  at  the  common  point 
of  intersection  of  these  three  lines. 

Let  M  (  Fig.  86)  be  the  middle  point 

of  BC,  N  the  middle  point  of  A  C,  and  G 

the  point  of  intersection  of  A  M  and  BN. 

By  comparison   of  the   similar    triangles 

ABC  and  NMC,  it  is  seen  that  NM  is 

equal  to  half  of  AB.     Comparing  the  similar  triangles  GAB  and 

GMNy  it  is  seen  that  GMis  half  of  A  G  and  GN  half  of  BG.    Hence 

GM  is  one-third  of  AM,  and  GN  one-third  of  BN. 

Any  plane  polygon,  regular  or  irregular,  may  be  subdivided  into 
triangles  whose  areas  and  centroids  may  be  computed  ;  so  that  the 
centroid  of  the  whole  area  can  be  determined  by  the  general  formulas 
of  Art.  154. 

Right  prism. — A  right  prism  may  be  generated  by  the  motion 
of  a  plane  area  perpendicular  to  itself.  The  centroid  of  the  area 
will  describe  a  line  which  must  contain  the  centroid  of  the  prism. 
For,  any  elementary  areas  will  generate  volumes  proportional  to 
the  areas  ;  and  if  moments  be  taken  with  respect  to  any  plane  con- 
taining the  line  generated  by  the  centroid  of  the  generating  area, 
the  moments  of  the  elementary  volumes  will  be  proportional  to 
the  moments  of  their  generating  areas.  But  the  moment  of  the 
whole  area  is  zero  for  such  a  plane  ;  hence  the  moment  of  the 
whole  volume  is  also  zero,  and  its  centroid  therefore  lies  in  the 
plane.  The  centroid  is  obviously  equidistant  from  the  two  bases 
of  the  prism. 


CENTROIDS. 


133 


Oblique  prism. —  An  oblique  prism  may  be  generated  by  the 
rectilinear  motion  of  a  plane  area  in  a  direction  not  perpendicular  to 
itself.  The  reasoning  given  for  the  case  of  a  right  prism  may  be 
applied  to  this  case,  with  a  like  result. 

Triangular  pyramid. — The  centroid  of  the  volume  of  a  triangular 
pyramid  lies  in  the  line  joining  the  vertex  with  the  centroid  of  the 
base.  To  prove  this,  let  A  be  the  vertex,  BCD  the  base,  and  Mthe 
centroid  of  the  base  (Fig.  87).  Let  B' CD'  be  a  plane  section*  par- 
allel to  BCD,  M'  being  its  centroid ;  it  may  be  proved  without 
difficulty  that  M'  lies  on  the  straight  line  AM.  Let  B" C" D"  be 
another  plane  section  parallel 
to  the  base,  and  let  a  prism  be 
generated  by  the  translation  of 
B'C'D'  parallel  to  AM  until 
it  falls  into  the  plane B" C" D" . 
The  centroid  of  this  prism  lies 
in  AM.  Let  any  number  of 
plane  sections  parallel  to  BCD 
be  taken,  and  let  each  be  the 
base  of  a  prism  generated  in 
the  same  way  as  the  one  de- 
scribed. Since  the  centroid  of 
each  prism  lies  upon  A  M,  so 
also  does  the  centroid  of  their 

combined  volume.  If  the  number  of  plane  sections  be  increased 
indefinitely,  their  distance  apart  approaching  zero  as  a  limit,  the 
volume  of  each  elementary  prism  approaches  zero,  but  their  com- 
bined volume  approaches  that  of  the  pyramid  ;  hence  the  centroid 
of  the  pyramid  lies  upon  AM. 

To  determine  the  position  of  the  centroid,  let  N  (Fig.  87)  be  the 
centroid  of  the  face  A  CD  ;  then  the  centroid  of  the  pyramid  must 
lie  on  BN,  and  must  therefore  be  the  point  of  intersection  of  AM 
and  BN.  If  E  is  the  middle  point  of  CD,  M  lies  upon  EB,  and 
EM=  EB/3;  also,  N  lies  upon  EA,  and  EN  =  EA/3.  The 
triangles   EAB    and   ENM   are    therefore    similar,  and   NM  = 


*  The  sections  B'C'D'  and  B"C"D"  are  not  shown  in  Fig.  87 ;  but  it  is 
to  be  understood  that  like  letters  denote  corresponding  vertices  of  the  three 
triangles  BCD,  B'C'D',  B"C"D". 


134  THEORETICAL    MECHANICS. 

AB/3.  Again,  comparing  the  similar  triangles  GAB,  GMN,  it 
is  seen  that 

GM=AG/3  =  AMJ^; 

GN  =  BG/3  =  BN/4. 

Therefore,  the  distance  of  the  centroid  from  the  base  is  equal  to 
one-fourth  the  altitude. 

Any  pyramid  or  cone. — The  above  proof  that  the  centroid  lies 
upon  the  line  drawn  from  the  vertex  to  the  centroid  of  the  base 
applies  to  a  pyramid  whose  base  is  any  polygon.  Moreover,  if  the 
base  be  divided  into  triangles,  the  pyramid  can  be  divided  into  tri- 
angular pyramids  whose  centroids  all  lie  in  a  plane  parallel  to  the 
base  and  at  a  distance  from  it  equal  to  one-fourth  the  altitude  ;  hence 
the  centroid  of  the  given  pyramid  also  lies  in  that  plane. 

If  the  base  is  bounded  by  any  curve,  this  may  be  regarded  as  the 
limit  of  an  inscribed  polygon  the  number  of  whose  sides  is  increased 
indefinitely.      Hence  the  following  proposition  may  be  stated  : 

The  centroid  of  any  pyramid  or  cone  lies  in  the  line  joining  the 
vertex  with  the  centroid  of  the  base,  a?id  at  a  distance  from  the  base 
equal  to  one-fourth  the  altitude. 

Examples. 

1.  Find  the  centroid  of  the  area  of  half  of  a  regular  hexagon. 

2.  Determine  the  centroid  of  an  area  composed  of  a  square,  and 
an  equilateral  triangle  one  of  whose  sides  coincides  with  a  side  of  the 
square. 

3.  Determine  the  centroid  of  a  volume  composed  of  a  cube,  and 
a  right  pyramid  whose  base  coincides  with  a  face  of  the  cube  and 
whose  altitude  is  equal  to  the  length  of  an  edge  of  the  cube. 

4.  Find  the  centroid  of  a  volume  made  up  of  a  right  circular 
cylinder  of  given  base  and  altitude,  and  a  cone  of  any  altitude  whose 
base  coincides  with  a  base  of  the  cylinder. 

5.  Prove  that  the  centroid  of  the  area  of  a  triangle  coincides  with 
the  mass- center  of  a  system  of  three  particles  of  equal  mass  situated 
at  the  vertices  of  the  triangle. 

6.  Determine  the  centroid  of  the  volume  of  a  frustum  of  a  cone 
or  pyramid. 

7.  Prove  that  the  centroid  of  the  volume  of  a  tetrahedron  coin- 
cides with  the  mass-center  of  a  system  of  four  particles  of  equal  mass 
situated  at  the  vertices  of  the  tetrahedron. 


CENTROIDS.  135 

§  3.  Determination  of  Centroids  by  Integration. 

159.  General  Formulas. —  In  many  cases  it  is  not  possible  to 
subdivide  the  body,  surface  or  line  whose  centroid  is  required  into 
finite  portions  whose  centroids  are  known.  In  such  cases,  if  an  exact 
result  is  desired,  resort  must  be  had  to  integration. 

Starting  with  the  general  formula 

x  =  (mxxx  +  m.x.,  +     .     .     :    )l(ml  -f  m2  +     .     .     .     ), 

let  the  number  of  parts  into  which  the  body  is  divided  be  increased, 
while  their  volumes  and  masses  become  smaller,  approaching  zero. 
Each  term  in  the  sum 

mxxy  +  m.,x.2  +     .     .     . 

approaches  zero,  while  the  sum  in  general  approaches  a  finite  limit, 
which  has  the  value 

limit  \mxXi  -j-  m%xt  -f     .     .     .     ]  —fxdM, 

the  limits  of  the  integration  being  so  taken  as  to  include  the  whole 
body.  The  denominator  in  the  expression  for  x  is  equal  to  M  (the 
whole  mass  of  the  body),  and  may  be  written  in  the  form/  dM.  The 
values  of  x,  y  and  z  may  therefore  be  written 

x=fxdMlfdM;    y  =fydAf/fdM;     z=fzdMjfdM.     (1) 

If  p  denotes  the  density  at  the  point  (x,  y>  z)  (being  in  the  most 
general  case  a  variable),  we  have  (Art.  153) 

dM=  PdV} 
and  the  equations  may  be  written 

x=fxpdVifpdV]    y=fypdV/fpdV; 

z^fzpdVjfpdV.  .  .         .     (2) 

Homogeneous  body. — If  the  body  is  homogeneous,  p  is  constant, 
and  the  formulas  reduce  to 

x  =fxdVlfdV;    y=fydV/fdV;     z=fzdV/fdV.     (3) 

These  are  also  the  formulas  for  the  coordinates  of  the  centroid  of  a 
volume. 

Any  area. — Starting  with  the  formulas  for  the  coordinates  of  the 
centroid  of  an  area  already  given  (Art.  155),  let  the  number  of  partial 
areas  be  increased  while  their  size  decreases,  approaching  zero  as  a 


136 


THEORETICAL    MECHANICS. 


limit.  The  values  of  x%  y  and  z  approach  limiting  values  identical  in 
form  with  those  given  by  equations  (3).  These  equations  may  there- 
fore be  used  in  finding  the  centroid  of  an  area  if  d  V  represents  an 
element  of  area.  In  case  of  a  plane  area,  only  two  of  the  equations 
are  needed,  the  plane  of  the  area  being  taken  as  the  plane  of  two  of 
the  coordinate  axes. 

Any  line. — The  values  of  x,  y  and  $  given  by  equations  (3)  may 
be  regarded  as  the  coordinates  of  the  centroid  of  any  line,  straight  or 
curved,  if  dV'is  taken  to  represent  an  element  of  length.  For  a 
plane  curve  only  two  of  the  equations  are  needed,  and  for  a  straight 
line  but  one  equation  is  needed.  The  centroid  of  a  straight  line  is 
evidently  at  its  middle  point. 

160.  Methods  of  Solution. —  In  applying  the  general  formulas 
for  center  of  mass,  center  of  volume,  center  of  area  and  center  of 
length  to  particular  problems,  many  special  methods  are  found  useful. 
Thus,  the  integration  may  be  double,  single,  or  triple,  depending  not 
only  upon  the  number  of  dimensions  of  the  body  considered,  but  also 
upon  the  way  in  which  the  differential  element  is  taken.  Again,  it 
is  often  convenient  to  employ  polar  coordinates,  or  to  otherwise 
change  the  variable  with  respect  to  which  the  integration  is  to  be 
performed.  These  various  devices  are  best  explained  in  connection 
with  the  problems  in  which  they  are  employed.  Several  such  prob- 
lems will  now  be  solved. 

161.  Applications. — I.  To  determine  the  centroid  of  a  circular  arc. 
Let  AB  (Fig.  88)  be  the  given 

arc,  the  radius  being  r.  By  sym- 
metry it  is  seen  that  the  centroid  lies 
upon  OX,  drawn  through  the  center 
bisecting  the  angle  A  OB.  Let  0  de- 
note the  angle  between  OX  and  the 
radius  vector  drawn  from  0  to  any 
point  of  the  arc.  If  s  represents  the 
length  of  the  arc,  measured  from 
some  fixed  point,  ds  replaces  dV  in 
the  general  formula  for  x ;  that  is, 

x  =fxdslfds. 
Introducing  0  as  the  variable, 

x  =  r  cos  0}     ds  =  r  d0.  Fig.  88. 


CENTROIDS. 


137 


If  the  angle  A  OB  is  a,  the  limits  of  integration  are  a/2  and 
Hence 


-a/2. 


r 


r2  cos  0<^0 


—a/ 2 


Xa/2 
-a/ 2 


rdd 


But 


x 


a/2 


•  /,«/2 

cos  0  </#  =  2  sin  a/2  ;     and     I        dd  =  a;     hence 

J — ali 

_  __  2r  sin  (a/2) r  sin  (a/2) 


a/2 


II.   To  find  the  centroid  of  the  area  of  a  triangle. 

Let  ABC  (Fig.  89)  be  the  triangle.     Draw  AX  perpendicular  to 
BC,  and  let  u  denote  the  length  of  B'C'}  drawn  parallel  to  BC  2X  a 
distance  x  from  A.     Let  a  denote  the  length  of  BC,  and  h  the  per- 
pendicular distance  from  A   to  BC; 
then  the  value  of  u  is  axjh.     In  the 
formula  x  —fxdVjfd V( Art.  159) 
there  may  be  substituted 

dV  =  udx  =  (ajh)xdx; 

and  therefore 

»4 


J 


dx 


x  = 


ih. 


£ 


x  dx 


Since  any  vertex  of  the  triangle  may 
be  used  instead  of  A,  the  above  result 
shows  that  the  ce?itroid  of  a  triangu- 
lar area  is  at  a  distance  from  either  side  equal  to  one-third  the  alti- 
tude measured  from  that  side. 

If  a  line  is  drawn  from  each  vertex  bisecting  the  opposite  side, 
the  three  lines  will  intersect  in  a  point  which  may  be  proved  geo- 
metrically to  coincide  with  the  centroid  as  above  determined.  (See 
Art.  158.) 

III.   To  find  the  centroid  of  the  area  of  a  circular  quadrant. 

Let  a  be  the  radius  of  the  circle,  and  let  OX  and  0  Y  (  Fig.  90) 
coincide  with  the  radii  bounding  the  quadrant.  Either  single  or 
double  integration  may  be  employed,  as  will  be  shown. 


138 


THEORETICAL    MECHANICS. 


Single  integration. — In  the  formula  for  x  the  element  of  area 
may  be  any  element,  all  parts  of  which  have  the  same  value  of  x. 

Hence  we  may  take 

dV '  =  y  dx, 
in  which  y  is  the  ordinate  of  the  curve 
at  the  point  whose  abscissa  is  x  ;  or, 
expressing  y  in  terms  of  x  by  means 
of  the  equation  of  the  circle, 
dV  =  {a2  —  x'1)1'1  dx 
Hence  the  formula  for  x  becomes 


fV  —  x2f'lxdx 

"  o 


Fig.  90. 


fo2  —  x^y'dx 

Reducing  numerator  and  denominator, 

f(a'  -  x2fxdx  ==  [-  i(«2  -  *•>*£  =  aV3  ; 
f(a2  —  x2Y!2dx  =  i[x(a'  —  x2)"2  +  a2  sin"1  (x/a)]" --=  ira2^; 

•/     O  ° 

x  =  (a"/^/(7ra2/4.)  =  \alyrr. 

From  symmetry  it  is  evident  that  y  =  x. 

Double  integration. — Let  the  element  of  area  be  taken  as 

dV  =  dx  dy. 


Then 


x  =Jfx  dx  dy  Iff  dx  dy  ;    y  =Jfy  dx  dy  Iff  dx  dy 


(  Notice  that  x  and  y  now  denote  the  coordinates  of  any  element 
dx  dy,  not  the  coordinates  of  the  curve.)  The  limits  of  the  integra- 
tions with  respect  to  the  two  variables  will  depend  upon  which  inte- 
gration is  performed  first.  If  the  ^-integral  is  first  taken,  the  limits 
for  x  are  o  and  V a 2  —  y  'l ;  since  for  any  value  of  y  these  are 
the  extreme  values  of  x  for  the  area  in  question.  The  integration 
with  respect  toy  would  then  have  the  limits  o  and  a.     Thus, 


x  = 


ff 

•So    *s  o 


xdy  dx 


dy  dx 


CENTROIDS.  I39 

Performing  the  ^-integration, 


O 

x  = 


1  C  Va*—y%<fy 


Taking  the  j-integral, 

*  =  («8/3)/(™V4)  =  4*/3w- 

The  value  of  y  may  be  found  in  a  similar  manner,  but  is  evidently 
equal  to  x. 

The  above  solutions  are  sufficient  to  illustrate  the  general  method 
to  be  adopted  in  determining  centroids  by  integration.  The  best 
choice  of  coordinates  and  of  elementary  volumes,  areas  or  lengths,  is 
a  matter  calling  for  mathematical  ingenuity.  The  matter  of  funda- 
mental importance  to  the  student  is  a  clear  understanding  of  the 
general  method,  rather  than  the  ability  to  solve  special  problems  by 
memory. 

Examples. 

1.  Determine  by  integration  the  centroid  of  the  volume  of  any 
pyramid  or  cone. 

2.  Determine  the  centroid  of  the  area  bounded  by  the  parabola 
y1  =  2px>  the  axis  of  x,  and  any  ordinate. 

3.  Find  the  centroid  of  the  area  bounded  by  the  parabola,  the 
axis  of  x,  and  any  two  ordi nates. 

4.  Find  the  centroid  of  the  volume  bounded  by  the  surface  of  a 
sphere  of  radius  a  and  two  parallel  planes  distant  xx  and  x.t  from  the 
center. 

Ans.  *=  [3^  +  ;r2)(2tf2  —  ^,2— ^/)]/[4(3^2— -^i2  —  xxx^— x22)\ 

5.  Find  the  centroid  of  the  area  of  any  zone  of  a  sphere. 

6.  Determine  the  centroid  of  the  area  bounded  by  a  circle  and 
any  two  parallel  lines. 

7.  From  a  circular  area  of  radius  r,  a  second  circular  area  of 
radius  r'  is  removed.  The  distance  between  the  centers  being  c, 
determine  the  centroid  of  the  area  remaining. 

8.  Determine  the  centroid  of  the  area  common  to  two  circles  of 
radii  r  and  r\  the  center  of  the  latter  lying  on  the  circumference  of 
the  former. 

9.  Determine  the  centroid  of  the  volume  of  a  hemisphere. 
Ans.  At  a  distance  from  the  center  equal  to  three-eighths  the 

radius. 


140  THEORETICAL    MECHANICS. 

10.  Determine  the  center  of  mass  of  a  hemisphere  in  which  the 
density  varies  directly  as  the  distance  from  the  center. 

An s.  At  a  distance  from  the  center  equal  to  two-fifths  the  radius. 

11.  Determine  the  center  of  mass  of  a  sphere  made  up  of  two 
hemispheres,  each  homogeneous,  but  one  twice  as  dense  as  the  other. 

12.  Determine  the  position  of  the  centroid  of  the  surface  of  a 
right  cone. 


CHAPTER   X. 

FORCES    IN    THREE    DIMENSIONS. 

§  i.   Concurrent  Forces. 

162.  Resultant  of  Any  Number  of  Concurrent  Forces.— The 

resultant  of  any  number  of  concurrent  forces,  whether  coplanar  or 
not,  may  be  found  by  the  repeated  application  of  the  principle  of  the 
parallelogram  or  triangle  of  forces.  The  resultant  of  any  two  is  a 
single  force  equal  to  their  vector  sum.  This  resultant,  combined 
with  a  third  force,  gives  as  the  resultant  of  -the  three  a  single  force 
equal  to  their  vector  sum  ;  and  so  on.  That  is,  the  resultant  of  any 
number  of  non-coplanar  concurrent  forces  is  a  single  force  equal  to 
their  vector  sum  and  acting  at  their  common  point  of  application. 

163.  Parallelopiped  of  Forces. —  Any  three  non-coplanar  con- 
current forces  and  their  resultant  may  be  represented  in  magnitude 
and  direction  by  three  edges  and  a 

diagonal  of  a  parallelopiped.  0[ D 

Thus,  let  OA,  OB  and  OC  (Fig.       A.-"  "A  /f7\ 

9 1 )   represent,  in  magnitude  and  di-         V        -f     \  ^ '      !     \ 
rection,    three    non-coplanar    forces.  \      /     ,'YB 

The  resultant  of  OA  and  OB  is  rep-  yij*^*^        >'   ■' 

resented  by  OC,  the  diagonal  of  the  Q  ^~C 

parallelogram  OA  C'B.     The  result-  Fig.  91. 

ant  of  OC  and  OC  is  represented  by 

OD,  the  diagonal  of  the  parallelogram  OC  DC.  But  OD  is  the 
diagonal  of  the  parallelopiped  of  which  OA,  OB  and  OC  are  adja- 
cent edges. 

164.  Resolution  of  a  Force  into  Three  Components. —  A  force 
may  be  resolved  into  three  components  acting  in  any  three  non- 
coplanar  lines  passing  through  a  point  in  its  line  of  action.  For  if 
the  given  force  be  represented  in  magnitude  and  direction  by  a  vector, 
a  parallelopiped  may  always  be  determined  of  which  this  vector  is  a 
diagonal,  and  whose  edges  are  parallel  to  the  three  chosen  lines. 

The  resolution  of  a  given  force  into  three  components  acting  in 
three  chosen  non-coplanar  lines  can  be  made  in  but  one  way  ;  for  a 
parallelopiped  is  completely  determined  if  a  diagonal  is  given  in 


142 


THEORETICAL    MECHANICS. 


length  and  direction,  and  if  the  directions  of  the  three  adjacent  edges 
are  also  given.* 

If  the  three  components  are  mutually  perpendicular,  their  mag- 
nitudes may  be  simply  computed  from  the  angles  they  make  with  the 
resultant.  Thus,  if  the  parallelopiped  in  Fig.  91  is  rectangular,  and 
if  a,  ft,  7  are  the  angles  between  OD  and  the  three  edges  OA,  OB, 
OC,  we  have 

OA  ==  OB  cos  a  ;     OB  =  OD  cos  /3 ;     OC  =  OD  cos  7. 
165.  Computation  of  Resultant  of  Concurrent  Forces. —  The 
simplest  method  of  determining  the  magnitude  and  direction  of  the 

resultant  of  non-coplanar  concurrent 
forces  is  usually  to  replace  each  force 
by  three  rectangular  components,  and 
then  combine  these  components. 

Thus,  having  given  any  concurrent 
forces,  of  magnitudes  Px,  Pti     .     .     .     , 
choose  a  set  of  rectangular  axes   OX, 
OY  and   OZ  (Fig.    92),    and   let  the 
angles  made  by  Px  with  these  three  axes 
respectively  be  a, ,  fix,  7, ,  with  similar 
notation  for  the  other  forces.    The  axial 
components  of  Px  are  Px  cos  c„  Px  cos  fix, 
Px  cos  yx ;  and  similar  expressions  may  be  written  for  the  compo- 
nents of  every  force.     The  whole  system  is  thus  reduced  to  three 
sets  of  collinear  forces,  which  combine  into  three  forces  as  follows  : 


Fig.  92. 


P.,  cos  a.,  -f- 


parallel  to  OX; 
"  "  OY; 
"        "   OZ. 


a  force  Px  cos  ax 

"      "     /\cos&-f />2cosft  + 

"      "     Px  cos  yx  +  />  cos  7,  + 

Let  R  denote  the  magnitude  of  the  resultant ;  a,  b,  c  its  angles  with 
OX,  OY,  OZ  ;  and  X,  Y,  Z  its  axial  components.     Then 

X  =  R  cos  a  =  Pi  cos  ax  -J-  P2  cos  a,  -f     .     .     .      ; 

Y  =  R  cos  b  =  Px  cos  /3x  +  P,  cos  /8a  -f     .     .     .      ; 

Z  =  R  cos  c  =  P,  cos  7j  +  P2  cos  y2  -j-     .     .     .      ; 

R*  =  X*  +  F2  +  Z2; 

Jf/^  ;     cos  0  =  F/Jfc  ;     cos  r  =  Z/7?. 


cos  « 


*  If  the  three  components  are  coplanar  with  the  given  force,  the  problem 
is  indeterminate.  If  only  two  of  the  components  are  coplanar  with  the  given 
force,  the  third  component  is  zero. 


FORCES    IN   THREE    DIMENSIONS.  1 43 

166.  Equilibrium. — The  general  condition  of  equilibrium  for  any 
system  of  concurrent  forces  is  that  the  resultant  is  zero,, 
From  the  equation 

R  =  V  X2  +  Y2  +  Z2  =  o 

may  be  derived  the  three  equations 

X  ==  o»     Y  =  o,     Z  =  0. 

For  if  these  three  equations  are  not  satisfied,  R  cannot  equal  zero 
unless  either  X2,  Y2  or  Z2  is  negative,  which  would  make  X,  Y  or 
Z  imaginary.  But  neither  of  these  quantities  can  be  imaginary  for 
any  real  system  of  forces. 

Since  the  directions  of  resolution  may  be  assumed  at  pleasure,  it 
follows  that 

If  a  system  of  concurrent  forces  is  in  equilibrium,  the  sum  of 
their  resolved  parts  in  any  direction  must  equal  zero. 

In  accordance  with  this  principle  an  infinite  number  of  equations 
may  be  written.  Only  three  such  equations  can  be  independent  as 
the  following  analysis  shows  : 

If  the  sum  of  the  resolved  parts  in  one  direction  is  zero,  the  re- 
sultant, if  not  zero,  must  act  in  a  plane  perpendicular  to  the  direction 
of  resolution.  If  the  sum  of  the  resolved  parts  is  zero  for  a  second 
direction,  the  resultant,  if  not  zero,  must  act  in  a  line  perpendicular  to 
the  two  directions  of  resolution.  If  the  sum  of  the  resolved  parts  is 
zero  for  a  third  direction  not  coplanar  with  the  other  two,  the  result- 
ant must  be  zero. 

It  follows  that  if  three  equations  be  formed  by  resolving  in  three 
non-coplanar  directions,  any  equation  obtained  by  resolving  in  a 
fourth  direction  may  be  derived  from  these  three. 

Examples. 

1.  A  particle  of  45  lbs.  mass  is  suspended  by  three  strings,  each 
8  ft.  long,  attached  at  points  A,  B,  Cm  the  same  horizontal  plane, 
the  distances  AB,  BC,  CA  being  each  4  ft.  Determine  the  tensions 
in  the  supporting  strings.  Ans.   16.37  l°s-  m  eacn  string. 

2.  A  particle  of  60  lbs.  mass  is  suspended  by  three  strings,  each 
12  ft.  long,  attached  at  points  A,  B,  C,  in  the  same  horizontal  plane, 
the  distance  AB  being  6  ft.  and  BC  and  CA  each  8  ft.  Determine 
the  tensions  in  the  supporting  strings. 

3.  A  particle  of  75  lbs.  mass  is  suspended  by  three  strings,  each 
10  ft.  long,  attached  at  points  A>  B,  C,  in  the  same  horizontal  plane, 


144  THEORETICAL    MECHANICS. 

the  distances  AB,  BC,  CA  being  4  ft.,  5  ft.  and  6  ft.  respectively. 
Required  the  tensions  in  the  supporting  strings. 

Ans.  33.62  lbs.,  9.04  lbs.,  35.84  lbs. 

4.  A  body  of  5  kilogr.  mass  is  suspended  by  a  string  which  is 
knotted  at  C  to  two  strings  CA,  CB,  these  being  attached  to  fixed 
supports  at  A  and  B  in  a  horizontal  plane.  At  C  is  knotted  a  fourth 
string  which  passes  over  a  smooth  peg  at  D  and  sustains  a  mass  of 
4  kilogr.  In  the  position  of  equilibrium  CD  is  horizontal  and  at 
right  angles  to  AB.  The  distances  A C,  BC}  AB  are  180  cm., 
150  cm.,  and  120  cm.  respectively.  Determine  the  position  of 
equilibrium  and  the  tensions  in  the  strings  AC,  BC 

Ans.  Inclination  of  plane  to  vertical,  38°  40'.  Tension  in  AC, 
1. 2 1  kilogr.;  tension  in  BC,  5.44  kilogr. 

§  2.    Composition  and  Resolution  of  Cotdples. 

167.  Representation  of  Couple  by  Vector. —  It  has  been  shown 
in  Chapter  IV  that  two  couples  applied  to  the  same  rigid  body  are 
equivalent  if  (1)  they  act  in  parallel  planes,  (2)  their  moments  are 
equal  in  magnitude,  and  (3)  their  rotation-directions  coincide.  So 
long  as  the  discussion  was  limited  to  coplanar  forces,  a  couple  could 
be  completely  represented  by  a  quantity  whose  numerical  magnitude 
specified  the  value  of  the  moment  and  whose  algebraic  sign  specified 
the  direction  of  rotation  in  the  plane.  It  is  now  needful  to  consider 
couples  whose  planes  are  not  parallel. 

A  couple  may  be  completely  represented  by  a  vector  in  the  follow- 
ing manner :  (a)  The  vector  is  to  be  taken  perpendicular  to  the 
plane  of  the  couple ;  (b)  its  length  must  rep- 
resent, to  some  chosen  scale,  the  magnitude 
of  the  moment  of  the  couple  ;  (c)  its  direction 
must  correspond  to  the  rotation-direction  of 
the  couple  in  its  plane.  These  requirements 
may  be  further  explained  as  follows : 

Let  the  plane  of  the  paper  be  parallel  to 

the  plane  of  the  couple ;    let  P  denote  the 

magnitude  of  each  force  and  p  the  length  of 

the  arm  ;  and  let  the  lines  of  action  and  direc- 

Fig.  93.  tions  be  as  shown  in  Fig.  93.     Then  {a)  the 

vector  which  represents  the  couple  must  be 

perpendicular  to  the  plane  of  the  paper  ;  (Jj)  its  length  must  be  Pp 

units  (the  unit  length  which  represents  one  unit  of  moment  being 


FORCES    IN   THREE    DIMENSIONS. 


145 


chosen  arbitrarily)  ;  (c)  it  must  be  decided  which  of  the  two  oppo- 
site directions  perpendicular  to  the  plane  of  the  paper  represents 
the  kind  of  rotation  shown  in  the  figure  and  which  the  opposite 
kind.  It  will  here  be  assumed  that  the  couple  shown  in  the  figure 
is  to  be  represented  by  a  vector  pointing  upward  from  the  paper. 
The  rule  thus  assumed  may  be  stated  generally  as  follows  : 

If  a  right-hand  screw  be  placed  with  its  axis  perpendicular  to  the 
plane  of  the  couple,  when  its  direction  of  rotation  coincides  with  that 
of  the  couple  its  direction  of  advance  will  coincide  with  the  direction 
of  the  vector  representing  the  couple. 

168.  Resultant  of  Any  Two  Couples. —  The  resultant  of  any 
two  couples  is  a  couple,  and  the  vector  representing  it  is  equal  to  the 
sum  of  the  vectors  representing  the  two  given  couples. 

This  has  already  been  proved  for  the  case  in  which  the  planes  of 
the  couples  are  parallel,  the  vector  sum  in  this  case  reducing  to  the 
algebraic  sum  (Art.  93). 

If  the  planes  of  the  two  couples  are  not  parallel,  let  the  plane  of 
the  paper  be  perpendicular  to  their  line  of  intersection,  this  line  being 
projected  at  B  (Fig.  94),  and  B M,  BN  being  the  traces  of  the  two 
planes.     Let    Gx  de- 
note the  moment  of 
the    couple     in     the 
plane  BM  and  G.2  the 
moment  of  the  couple 
in  the  plane  BN,  their 
directions  of  rotation 
being  such  that  they 
are  represented  by  the 
vectors  A'B\  B'C. 

Replace  the  for- 
mer couple  by  an 
equivalent  couple  of 
which  the  forces  are 

perpendicular  to  the  plane  of  the  paper,  P  being  the  magnitude  of 
each  force  and  Gl/P  its  arm.  One  of  these  forces  may  be  taken  as 
acting  at  B  and  the  other  at  A,  if  AB  =  GX\P.  Similarly,  the 
second  couple  may  be  replaced  by  a  couple  of  forces  P,  one  act- 
ing at  B,  opposite  to  the  force  P  already  assumed  to  act  at  B,  and 
the  other  at  C,  if  BC  =  G.JP.  Since  the  two  forces  applied  at  B 
10 


Fig.  94. 


146  THEORETICAL    MECHANICS. 

balance  each  other,  the  two  couples  are  equivalent  to  a  couple  of 
equal  and  opposite  forces  P  applied  at  A  and  C,  its  moment  being 
PX  AC 

It  may  be  shown  that  this  resultant  couple  is  completely  repre- 
sented by  the  vector  A'C.  In  the  triangles  A'B'C,  ABC,  the 
side  A' B'  is  perpendicular  to  AB  and  equal  to  P  X  AB  ;  and  the 
side  B'C  is  perpendicular  to  BC  and  equal  to  P  X  BC.  There- 
fore A'C  is  perpendicular  to  AC  and  equal  to  P  X  AC>  which  is 
the  moment  of  the  resultant  couple.  It  is  also  evident  that  the  direc- 
tion of  the  vector  A  '  C  represents  correctly  the  rotation-direction  of 
the  resultant  couple.     Thus  the  proposition  is  proved. 

169.  Resultant  of  Any  Number  of  Couples;  Resolution  of  a 
Couple. —  By  repeated  applications  of  the  above  principle,  any  num- 
ber of  non-coplanar  couples  may  be  combined  ;  it  is  seen  that  the 
resultant  will  always  be  a  couple,  and  that  the  vector  representing 
it  is  the  sum  of  the  vectors  representing  the  several  couples.  This 
may  be  briefly  expressed  by  saying  that  the  resultant  couple  is  the 
vector  sum  of  the  given  couples. 

By  reversing  this  process,  a  given  couple  may  be  replaced  by 
several  component  couples. 

Since  in  general  a  vector  may  be  resolved,  in  one  way  only,* 
into  three  components  whose  directions  are  given,  a  couple  may  be 
resolved  (in  one  way  only)  into  three  component  couples  whose 
planes  are  given. 

The  simplest  method  of  computing  the  resultant  of  any  number 
of  couples  is  usually  to  replace  each  by  its  components  parallel  to 
three  rectangular  planes.  The  whole  system  is  thus  reduced  to  three 
sets  of  coplanar  couples,  each  of  which  reduces  to  a  single  couple  by 
algebraic  addition  of  the  several  moments.  IfZ,  My  iVare  the  mo- 
ments of  these  partial  resultants,  the  final  resultant  has  the  moment 


G  =  VLi  +  M'1  +  N\ 

If  /,  m,  n  are  the  angles  between  the  vector  representing  the  resultant 
couple  and  the  vectors  representing  its  three  components, 

cos  /  =  L/G  ;    cos  m  =  M/G  ;    cos  n  =  NjG. 

*  If  the  three  components  are  coplanar,  the  resolution  is  impossible  un- 
less the  given  vector  lies  in  the  plane  of  the  components  ;  in  which  case  the 
resolution  can  be  made  in  an  infinite  number  of  ways. 


FORCES    IN    THREE    DIMENSIONS. 


§  3.  Any  System  of  Forces. 


147 


170.  Any  System  Reduced  to  a  Force  and  a  Couple. —  Any 

system  of  forces  applied  to  a  rigid  body  is  equivalent  to  a  single  force 
and  a  couple. 

Let  P  represent  the  magnitude  of  any  one  of  the  given  forces. 
At  any  point  O,  introduce  a  force  equal  and  parallel  to  P  and  an 
equal  and  opposite  force.  One  of  these,  with  the  given  force,  forms 
a  couple ;  so  that  the  given  force  is  equivalent  to  an  equal  force  P 
applied  at  O  and  a  couple.  In  a  similar  manner  each  one  of  the 
given  forces  may  be  replaced  by  an  equal  force  applied  at  O  and  a 
couple.  The  whole  system  is  therefore  equivalent  to  a  system  of 
concurrent  forces  applied  at  O,  equal  in  magnitude  and  direction  to 
the  given  forces,  together  with  a  system  of  couples.  The  former 
combine  into  a  single  force  equal  to  the  vector  sum  of  the  given 
forces  ;  and  the  latter  into  a  single  couple,  equal  to  the  vector  sum 
of  the  several  couples. 

171.  Computation  of  the  Force  and  Couple. —  The  single 
force  which  results  from  the  above  process  of  combination  may  be 
computed  as  if  the  forces  were  concurrent  (Art.  165);  its  magnitude 
and  direction  are  the  same,  whatever  point  is  chosen  as  its  point  of 
application.  The  couple  may  be  computed  most  readily  by  a  differ- 
ent method  of  resolution. 

Choosing  a  set  of  rectangular  axes,  OX,  0  Fand  OZ,  let  x, y}  z  be 
the  coordinates  of  the  point  of  application  of  any  force  P,  and  let  P 
be  replaced  by  its  axial  components  X,  Fand  Z  (Fig.  95).  Each 
of  these  components  may  be  replaced  by  an  equal  force  at  O  and 
two  couples,  as  follows  :  Consider  the  force  X applied  a.tA(  Fig.  95). 
Introduce,  in  the  line  CB,  a  force  equal  to  X  and  an  equal  and  oppo- 
site force  ;  and  in  the  line  OX  also  a  pair  of  equal  and  opposite 
forces  of  magnitude  X.  The  force  X  at  A  and  the  equal  and  oppo- 
site force  in  the  line  CB  form  a  couple  whose  plane  is  parallel  to 
the  plane  xy  and  whose  moment  *  is  — Xy.  The  remaining  force 
X  in  the  line  CB  and  the  equal  and  opposite  force  in  the  line  OX 

*  The  moments  of  couples  in  planes  parallel  to  the  yz  plane  are  called 
plus  or  minus,  according  as  the  vectors  representing  them  point  toward  the 
plus  or  the  minus  direction  of  the  .r-axis.  A  similar  convention  is  adopted 
for  each  of  the  other  coordinate  planes. 


148 


THEORETICAL    MECHANICS. 


form  a  couple  lying  in  the  plane  zx,  of  moment  Xz.  These  two 
couples,  with  the  remaining  force  in  the  line  OX,  are  equivalent  to 
the  given  force  X  applied  at  A. 

In  a  similar  manner  it  may  be  shown  that  the  force  Y  (Fig.  95) 
is  equivalent  to  an  equal  force  applied  at  0;  a  couple  parallel  to  the 
plane  yz,  of  moment  —  Yz  ;  and  a  couple  parallel  to  the  plane  xy,  of 
moment  Yx.  Also,  Z  (Fig.  95)  is  equivalent  to  an  equal  force  ap- 
plied at  0;  a  couple  parallel  to  the  plane  zx,  of  moment  — Zx;  and 
a  couple  parallel  to  the  plane  yz,  of  moment  Zy.  * 

Combining  the  three  forces  at  0  into  a  single  force,  and  the  six 
couples  into  three,  it  is  seen  that  the  given  force  P,  applied  at  A, 


is  equivalent  to  an  equal  and  parallel  force  applied  at  O,  together 
with  three  couples  as  follows  : 

In  the  ^-plane,  a  couple  of  moment  Zy  —  Yz. 

In  the  zx- plane,  a  couple  of  moment  Xz  —  Zx. 

In  the  .zy-plane,  a  couple  of  moment  Yx  —  Xy. 

These  three  couples  may  be  represented  by  vectors  having  the 
directions  OX,  0  Y  and  OZ,  respectively,  and  of  lengths  represent- 
ing the  values  of  the  moments. 

Let  there  be  given  any  number  of  forces,  Px  applied  at  the  point 


*  These  results  may  be  obtained  without  actually  repeating  the  process 
employed  in  case  of  the  force  X,  by  cyclic  substitution  of  the  letters  xyz  and 


XYZ. 


FORCES    IN    THREE    DIMENSIONS.  1 49 

C*ii  y\y  #i)»  ^2  applied  at  the  point  (x2,  y2,  z2),  etc. ;  and  let  each  be 
replaced  by  a  force  and  three  couples  in  the  manner  described.  Let 
Lx  =  Zxyx  —  Yxzx ;  Mx  =  Xxzx  ~  Zxxx  ;  Nx  =  Yxxx  -  Xxyx ; 
with  similar  notation  for  each  force.  The  whole  system  is  equivalent 
to  a  force  R,  applied  at  O,  equal  to  the  vector  sum  of  the  given 
forces  ;  and  a  couple  which  is  the  resultant  of  three  couples  as  fol- 
lows : 

In  the  yz-p\ane,  a  couple  of  moment  Lx  -f-   Z2  -f-    .    .    .   =  L. 

In  the  gar-plane,  a  couple  of  moment  Mx  +  M2  -f-    .    .    .   =  M. 

In  the  ^/-plane,  a  couple  of  moment  Nx  -f  N2  ~\-  .  .  .  =  N. 
The  moment  of  the  resultant  couple  is 


G  =  V  D  +  M*  +  N2; 
and  the  direction  of  its  plane  may  be  found  as  in  Art.  169. 

172.  Resultant  of  any  System  of  Forces. —  Having  reduced 
a  system  of  forces  to  a  force  and  couple  as  above,  if  it  is  attempted 
to  reduce  it  to  a  still  simpler  system,  it  appears  that  in  certain  cases 
it  may  be  reduced  to  a  couple  and  in  other  cases  to  a  single  force  ; 
but  that  in  general  the  simplest  system  to  which  it  can  be  reduced 
consists  of  two  non-coplanar  forces. 

(a)  If  the  vector  sum  of  the  given  forces  is  o,  the  resultant  is 
evidently  a  couple,  determined  as  above. 

(b)  If  the  plane  of  the  couple  found  by  the  above  process  is  par- 
allel to  the  single  force  R,  this  force  and  couple  are  equivalent  to  a 
single  force  (Art.  94)  which  is  the  resultant  of  the  given  system. 
This  resultant  force  is  equal  to  the  vector  sum  of  the  given  forces, 
and  its  line  of  action  can  be  determined  as  in  Art.  94. 

(V)  If  the  plane  of  the  couple  is  not  parallel  to  the  force  R,  let  it 
be  replaced  by  an  equivalent  couple  of  which  one  force  acts  in  a  line 
intersecting  the  line  of  action  of  R.  This  force  may  be  combined 
with  R,  thus  reducing  the  system  to  two  non-coplanar  forces.  This 
reduction  to  two  forces  may  be  made  in  an  infinite  number  of  ways, 
but  in  general  no  further  reduction  is  possible. 

§  4.  Equilibrium. 

173.  Equations  of  Equilibrium. — Let  a  system  of  forces  ap- 
plied to  the  same  rigid  body  be  reduced  to  a  force  and  couple  as  in 
Art.  171.     Let  R  denote  the  magnitude  of  the  force,  a,  b,  c  the 


I50  THEORETICAL    MECHANICS. 

angles  it  makes  with  the  coordinate  axes,  and  X,  Y,  Z  its  axial 
components  ;  and  let  G  denote  the  moment  of  the  couple,  /,  mt  n 
the  angles  made  with  the  axes  by  the  vector  representing  the  couple, 
and  L,  M,  N  the  moments  of  the  three  component  couples  parallel 
to  the  coordinate  planes. 

In  order  that  the  system  may  be  in  equilibrium,  both  R  and  G 
must  reduce  to  zero  ;  otherwise  the  force  and  couple  may  be  reduced 
(as  in  Art.  172)  either  to  a  single  force,  to  a  couple,  or  to  two  non- 
coplanar  forces. 

The  condition  R  =  o  requires  that 

X=o;      Y=o;     Z  =  o;      .         .  .      (1) 

and  the  condition  G  =  o  requires  that 

L  =  o  ;     M  =  o  ;     N  =  o.     .  .         .     (2) 

For  unless  X,  Y  and  Z  are  severally  equal  to  zero,  either  X2,  Y2  or 
Z2  must  be  negative,  and  X,  Y  or  Z  therefore  imaginary.  But  from 
the  manner  of  computing  these  quantities  it  is  evident  that  they  must 
be  real.     Similar  reasoning  applies  to  L,  M  and  N. 

Equations  (1)  and  (2)  are  therefore  six  independent  equations  of 
equilibrium.  Since  the  position  of  the  origin  may  be  chosen  at 
pleasure,  equations  (2)  maybe  written  in  an  infinite  number  of  ways. 
Also,  since  the  directions  of  the  axes  may  be  chosen  arbitrarily,  equa- 
tions (1)  may  be  written  in  an  infinite  number  of  ways. 

In  order  that  the  conditions  of  equilibrium  may  be  stated  in  a 
concise  form,  it  is  convenient  to  generalize  the  definition  of  moment 
of  a  force  in  the  following  manner. 

174.  Moment  of  a  Force  About  an  Axis. —  The  moment  of  a 
force  with  respect  to  an  axis  has  been  defined,  for  the  case  in  which 
the  axis  is  perpendicular  to  a  plane  containing 
the  line  of  action  of  the  force  (Art.  70),  as 
the  product  of  the  magnitude  of  the  force  into 
the  perpendicular  distance  of  its  line  of  action 
from  the  axis.  Thus,  if  the  line  of  action  of 
the  force  P  lies  in  the  plane  of  the  paper 
(Fig.  96)  and  the  axis  is  perpendicular  to 
that  plane  at  O,  the  moment  of  the  force  is  equal  to  Pp. 

If  the  axis  is  not  perpendicular  to  the  force  (that  is,  to  any  plane 
containing  the  line  of  action  of  the  force),  the  moment  is  to  be  com- 


FORCES    IN    THREE    DIMENSIONS.  151 

puted  by  replacing  the  force  by  two  components,  one  parallel  and 
the  other  perpendicular  to  the  axis,  and  taking  the  moment  of  the 
latter. 

175.  Conditions  of  Equilibrium. —  Referring  to  the  process  by 
which  a  system  is  reduced  to  a  single  force  and  a  couple  (Art.  171), 
it  is  seen  that  Zxyx  —  Yxzx  is  equal  to  the  moment  of  the  force  Px 
with  respect  to  the  axis  OX;  Xxzx  —  ZXXX  to  its  moment  with  re- 
spect to  O  Y;  and  Yxxx  —  Xxyx  to  its  moment  with  respect  to  OZ. 
The  quantity  L  is  therefore  equal  to  the  sum  of  the  moments  of  the 
given  forces  about  the  axis  OX ;  M  to  the  sum  of  their  moments 
about  O  Y;  and  N  to  the  sum  of  their  moments  about  OZ. 

The  meaning  of  the  two  sets  of  equations  of  equilibrium  (Art. 
173)  may  therefore  be  stated  in  words  as  follows  : 

(1)  The  sum  of  the  resolved  parts  of  the  given  forces  parallel  to 
each  of  the  axes  is  zero. 

(2)  The  sum  of  the  moments  of  the  given  forces  about  each  of 
the  given  axes  is  zero. 

Since  any  line  whatever  may  be  taken  as  one  of  the  axes,  it  fol- 
lows that,  for  equilibrium, 

(a)  The  sum  of  the  resolved  parts  of  the  forces  in  any  direction 
is  zero. 

(b)  The  sum  of  their  moments  about  any  axis  is  zero. 

These  principles  lead  to  an  infinite  number  of  equations  ;  but  only 
six  can  be  independent. 

Examples. 

1.  A  three-legged  stool  rests  upon  a  smooth  horizontal  floor.  De- 
termine the  pressures  at  the  three  points  of  support. 

2.  The  three  points  of  support  of  a  three-legged  stool  being  at 
any  distances  apart,  what  must  be  the  position  of  its  center  of  gravity 
in  order  that  the  pressures  at  the  points  of  support  may  be  equal? 

3.  A  four-legged  table  rests  upon  a  smooth  horizontal  floor.  How 
much  can  be  determined  about  the  supporting  forces  ? 

4.  A  windlass  consists  of  a  circular  cylinder  supported  at  two 
points  in  bearings  which  permit  free  revolution  about  its  axis,  a  crank 
attached  to  one  end  of  the  cylinder,  and  a  cord  wound  upon  the 
cylinder  and  having  one  end  free.  To  the  free  end  of  the  cord  is 
attached  a  heavy  body  which  is  to  be  lifted  by  the  application  of  a 
force  to  the  crank-handle.  The  axis  of  revolution  of  the  cylinder  is 
horizontal.  Assuming  that  the  force  applied  to  the  crank  is  perpen- 
dicular both  to  the  axis  of  revolution  and  to  the  crank-arm,  and  that 


152 


THEORETICAL    MECHANICS. 


the  bearings  are  frictionless,  it  is  required  to  determine  all  forces 
acting  on  the  windlass  when  the  weight  of  the  lifted  body  is  known. 
Let  W  —  weight  of  body  lifted  ;  P  =  force  applied  to  crank- 
handle  ;  r=  radius  of  cylinder;  a  =  length  of  crank-arm  (z.  e., 
radius  of  circle  described  by  point  of  application  of  P) ;  /  =  distance 
between  centers  of  bearings ;  b  =  distance  from  center  of  circle  de- 
scribed by  point  of  application  of  P  to  nearest  bearing ;  lx  and 
/.2  =  distances  from  centers  of  bearings  to  suspended  cord.  The 
bearings  being  assumed  smooth,  the  supporting  forces  exerted  by 
them  may  have  any  directions  perpendicular  to  the  axis  of  the  cylin- 
der ;  let  them  be  replaced  by  horizontal  and  vertical  components,  and 
let  Hx ,  Vx  denote  the  components  of  the  force  at  the  bearing  near 
the  crank  and  H2 ,  Vt  the  components  of  the  force  at  the  other  bear- 
ing.    The  direction  of  the  force  Pt  if  always  perpendicular  to  the 


Fig.  99. 


crank-arm,  will  vary   as    the    cylinder  revolves.       Let    6  =  angle 
between  crank-arm  and  the  vertical. 

All  forces  acting  upon  the  windlass  act  in  planes  perpendicular  to 
the  axis  of  revolution ;  hence,  if  forces  are  resolved  parallel  to  this 
axis,  no  equation  results.     Five  independent  statical  equations  may, 
however,  be  written  as  follows  : 
Resolving  horizontally, 

Hx  +  H%  +  Pcos  0  =  o.        .         .  .     (1) 

Resolving  vertically, 

V1  +   V2  —  Psin  6—  W=  o.         .  .     (2) 

Taking  moments  about  the  axis  of  revolution, 

Pa  —  Wr  =  o (3) 

Taking  moments   about    a   vertical    line    through    center  of    circle 
described  by  point  of  application  of  P, 

Hxb  +  H.t{b  +  /)  =  o.        .         .         .     (4) 


FORCES    IN    THREE    DIMENSIONS.  1 53 

Taking-  moments  about  a  horizontal  diameter  of  same  circle, 

Vxb  +  Vlb  +  /)  -  W{b  +  /,)  =  o.     .        .     (5) 

These  five  equations  contain  five  unknown  quantities,  Hx ,  Vx ,  H.t , 

V2 ,  P.     Their  solution  is  simple.     Thus,  (3)  determines  P  at  once  ; 

(1)  and  (4)  then  determine  Hx  and  H.z ;  (2)  and  (5)  determine  Vx 

and  V,. 

5.  In  Ex.  4,  take  numerical  data  as  follows  :  lx  =  7/3 ;  b  = 
//io  ;   #  =  6r.     Solve  for  0  =  o,  90°,  180°,  270°,  45°. 

^tzj.  P  =  W/6  for  any  value  of  6.  For  #  =  o,  //,  = 
-(u/6o)W,  #,  =  (1/60)^,  ^  =  (2/3)^,  Vt  =  (1/3)^ 
For  0  =  90°,  Hx=  H2  =  o,  Ft  -  (51/60)  W,  r2  = 
(19/60)^. 

6.  In  Ex.  4,  substitute  for  the  force  P  a  weight  Q  suspended  from 
the  crank-handle.  If  Q  is  known,  determine  the  position  of  equi- 
librium and  all  unknown  forces. 

7.  Assume  dimensions  as  in  Ex.  5,  and  let  the  weight  Q,  applied 
as  in  Ex.  6,  be  equal  to  W.  Determine  position  of  equilibrium  and 
all  forces. 

Ans.  0  =  90  36'  or  1700  24' ;  Hx  =  Ht  =  o ;  Vx  =  (17/20)  W; 
V,  =  (19/60)  w. 

8.  A  circular  cylinder  is  placed  with  axis  horizontal,  one  end  rest- 
ing in  a  smooth  cylindrical  bearing  and  the  other  on  an  inclined 
timber.  The  longitudinal  axis  of  the  timber  lies  in  a  plane  perpen- 
dicular to  the  axis  of  the  cylinder  and  is  inclined  to  the  horizontal  at 
a  known  angle.  If  its  surface  is  so  rough  as  to  prevent  sliding,  can 
the  cylinder  be  in  equilibrium  ? 

9.  In  Ex.  8,  can  equilibrium  be  produced  by  hanging  a  weight 
from  a  string  wound  on  the  surface  of  the  cylinder,  assuming  sliding 
to  be  impossible?  If  equilibrium  is  thus  possible,  specify  the  re- 
quired weight,  and  all  forces  acting  on  the  cylinder. 

10.  Two  timbers  are  placed  with  longitudinal  axes  parallel  and 
with  upper  surfaces  in  the  same  horizontal  plane.  A  circular  cylin- 
der rests  upon  the  timbers.  Can  the  supporting  timbers  be  tilted 
without  destroying  the  equilibrium  of*  the  cylinder,  assuming  the  sur- 
faces to  be  sufficiently  rough  to  prevent  sliding  ? 

1 1.  The  points  of  support  of  a  three-legged  stool  form  an  equi- 
lateral triangle  ABC  oi  side  a.  The  center  of  gravity  is  vertically 
above  the  centroid  of  the  triangle.  The  stool  is  pulled  horizontally 
by  a  string  lying  in  the  vertical  plane  containing  AB  and  at  a  dis- 
tance //  above  the  floor.  Show  that  the  stool  will  tip  without  sliding 
if  the  coefficient  of  friction  is  greater  than  a^h. 

12.  A  uniform  straight  bar  is  placed  with  one  end  on  a  rough 
horizontal  floor  and  the  other  in  the  right  angle  between  two  smooth 
vertical  walls.     Show  that  it  will  be  in  equilibrium  if  its  inclination 


154  '  THEORETICAL    MECHANICS. 

to   the   vertical  is    less  than  tan~'  (2/x),  /j,  being  the   coefficient  of 
friction. 

13.  In  Ex.  12,  show  that  the  resultant  force  acting  upon  the  bar 
at  either  end  acts  in  the  vertical  plane  containing  the  bar.  Deter- 
mine these  forces. 

14.  A  bar  is  placed  with  one  end  upon  a  rough  horizontal  floor 
and  the  other  against  a  rough  vertical  wall.  Determine  the  posi- 
tions of  limiting  equilibrium. 


CHAPTER   XI. 

GRAVITATION. 

§  i.  Attraction  Between   Two  Particles. 

176.  Law  of  Gravitation. —  Every  portion  of  matter  exerts  an 
attractive  force  upon  every  other  portion.  The  magnitude  and  direc- 
tion of  this  attractive  force  for  any  two  bodies  may  be  determined  in 
accordance  with  Newton's  law  of  universal  gravitation,  which  may  be 
stated  as  follows  : 

Every  particle  of  matter  attracts  every  other  particle  with  a  force 
which  acts  along  the  line  joining  the  two  particles,  and  whose  magni- 
tude is  proportional  directly  to  the  product  of  their  masses  and 
inversely  to  the  square  of  the  distance  between  them. 

The  proportionality  expressed  in  this  law  may  be  stated  algebra- 
ically as  follows  :  Let  P  denote  the  magnitude  of  the  attractive  force 
between  two  particles  whose  masses  are  mx,  m.z,  and  whose  distance 
apart  is  r ;  and  P'  the  force  between  two  particles  whose  masses  are 
ml,  m.2\  and  whose  distance  apart  is  r' .     Then 

PIP'  =  (nixmj  r2)l{mlm.;  I  r'*). 

This  equation  may  be  written  in  the  form 

Pr2lmxm2  =  P'r^/m^ml  =  7. 

Since  a  similar  equation  may  be  written  for  any  pair  of  particles 
whatever,  the  law  of  gravitation  may  be  expressed  by  the  equation 

Prillmxm.l  =  7, 
or 

P  =  rpnjnjr*, 
in  which  7  is  a  constant. 

The  value  of  7  is  to  be  determined  by  experiment ;  but  having 
been  determined  for  one  case  it  is  known  for  all  cases. 

This  formula  applies  strictly  only  to  particles,  but  it  gives,  to  a 
close  approximation,  the  attraction  between  two  bodies  of  finite  size 
whose  linear  dimensions  are  small  compared  with  the  distance  be- 
tween them.     We  shall  first  consider  the  attraction  between  two 


156  THEORETICAL    MECHANICS. 

particles,  and  shall  then  give  a  brief  discussion  of  methods  of  com- 
puting accurately  the  attraction  between  bodies  of  finite  size. 

Attraction  is  a  mutual  action  between  two  particles  or  bodies  ; 
i.  e. ,  each  exerts  an  attractive  force  upon  the  other,  the  two  forces 
being  equal  in  magnitude  and  opposite  in  direction.  This  is  implied 
in  the  above  statement  of  the  law  of  gravitation.  It  is  also  in  ac- 
cordance with  the  law  of  "action  and  reaction,"  Newton's  third  law 
of  motion  (Art.  35). 

177.  The  Constant  of  Gravitation.  —  The  quantity  7  in  the 
above  formula  is  called  the  constant  of  gravitation.  Its  numerical 
value  depends  upon  the  units  in  which  force,  mass  and  distance  are 
expressed.  For  any  given  system  of  units,  7  is  numerically  equal  to 
the  attractive  force  between  two  particles  of  unit  mass  at  unit  dis- 
tance apart  ;  for  if  mx  =  I,  m%  =  1  and  r  =  I,  the  formula  gives 

If  mass,  length  and  force  be  expressed  in  any  units  employed  in 
ordinary  practical  problems,  the  numerical  value  of  7  is  very  small. 
Thus  if  mass  is  expressed  in  pounds  and  distance  in  feet,  7  is  equal 
to  the  attractive  force  between  two  particles  of  one  pound  mass  each, 
placed  one  foot  apart.  This  force  is  too  small  to  be  detected  by  ordi- 
nary methods  of  measuring  forces  ;  and  if  expressed  in  terms  of  any 
unit  in  common  use  its  numerical  value  is  very  small.  (See  Arts. 
184,  186.) 

178.  Gravitation  Unit  of  Mass. —  Instead  of  choosing  the  units 
of  mass,  length  and  force  independently,  they  may  be  so  chosen  as 
to  give  7  any  desired  value.  Moreover,  two  of  these  units  may  still 
be  chosen  arbitrarily,  the  third  being  then  so  taken  as  to  satisfy  the 
equation  P  =  ym^mjr2  with  the  desired  value  of  7. 

In  order  to  simplify  the  equation,  let  7  =  1.  Then  the  units 
must  be  so  chosen  that  two  particles,  each  of  unit  mass,  placed  at  the 
unit  distance  apart,  attract  each  other  with  unit  force.  The  unit  mass 
which  satisfies  this  condition,  the  other  units  having  been  chosen  at 
pleasure,  is  called  the  gravitation  unit. 

To  determine  the  value  of  this  unit  of  mass  in  terms  of  the  pound, 
gram,  or  other  known  unit,  requires  the  same  process  of  experiment 
and  reasoning  which  is  involved  in  the  determination  of  the  constant 
7  when  all  the  units  have  been  chosen.  This  subject  will  be  resumed 
in  Art.  185. 


GRAVITATION. 


Examples. 


157 


1.  If  two  particles  20  ft.  apart  attract  each  other  with  a  force  of 
12  lbs.,  with  what  force  will  they  attract  each  other  if  placed  50  ft. 
apart  ? 

2.  If  two  particles  of  1 ,  000  grams  and  1 2, 000  grams  mass  respect- 
ively attract  each  other  with  a  force  equal  to  the  weight  of  P  grams 
when  30  cm.  apart,  with  what  force  will  two  particles  of  1,800  grams 
and  3,000  grams  attract  each  other  when  100  cm.  apart? 

Ans.  o.  0405P  grams  weight. 

3.  In  Ex.  2,  what  is  the  value  of  the  constant  7  in  terms  of  P? 

Ans.  7  =  3/740,000. 

4.  With  the  data  of  Ex.  (2),  what  is  the  gravitation  unit  of  mass? 

Ans.  A  mass  equal  to  200/1/(3/*)  grams. 
[The  answers  given  to  examples  3  and  4  imply  that  the  centi- 
meter is  the  unit  length  and  the  weight  of  a  gram  the  unit  force.] 

179.  Dimensions  of  Units. —  The  formula  expressing  the  law  of 
gravitation  for  two  particles  may  be  written  Pr2lm1m2  =  7.  The 
constant  7  is  therefore  of  dimensions  (Art.  14) 

PL'/M1 

if  the  units  of  force,  length  and  mass  are  all  taken  as  fundamental, — 
i.  e.,  are  all  chosen  independently  (Art.  13). 

The  law  of  gravitation,  however,  itself  furnishes  a  means  of  mak- 
ing one  of  these  units  depend  upon  the  other  two.  For  the  equa- 
tion expressing  the  law  may  be  written 

P  =  Mjfftjr*, 

if  the  units  be  so  chosen  as  to  satisfy  the  condition  stated  in  Art.  178. 
The  relation  between  the  units  will  then  be  expressed  by  the  dimen- 
sional equation,  p  __  M2/L2 

Any  two  of  these  units  being  chosen  arbitrarily,  the  dimensions  of 
the  third  are  given  by  this  equation. 

Thus,  if,  as  in  Art.  178,  the  units  of  force  and  length  are  made 
fundamental,  the  dimensions  of  the  unit  mass  are  expressed  by  the 
equation  M  =  LF„2 

The  relation  among  units  which  is  expressed  by  a  dimensional 
equation  is  thus,  to  a  certain  extent  at  least,  arbitrary.  This  will  be 
further  illustrated  when  the  kinetic  or  ' '  absolute ' '  system  of  units  is 
explained.      (See  Art.  219.) 


158  THEORETICAL    MECHANICS. 

§  2.  Attractions  of  Spheres  and  of  Spherical  Shells. 

180.  Law  of  Gravitation  Applied  to  Continuous  Bodies. — The 

law  of  attraction  is  stated  above  as  applying  to  particles,  and 
these  are  treated  as  bodies  of  finite  mass  but  without  finite  size. 
If  a  body  consisted  of  a  finite  number  of  such  particles,  its  resultant 
attraction  for  any  other  body  or  particle  would  be  computed  by  find- 
ing the  resultant  of  the  attractive  forces  due  to  all  its  individual  par- 
ticles. 

If,  instead  of  being  made  up  of  discrete  particles  of  finite  mass,  a 
body  occupies  space  continuously  (Art.  5),  so  that  any  portion  whose 
mass  is  finite  is  of  finite  volume,  the  resultant  attraction  Of  one  body 
upon  another  may  be  computed  in  a  similar  manner,  but  the  process 
involves  integration. 

Let  mx  and  m.L  denote  the  total  masses  of  two  continuous  bodies, 
and  Amx ,  Am,  any  small  elements  of  these  masses.  If  the  dimensions 
of  these  elementary  portions  are  small  in  comparison  with  their  dis- 
tance apart,  the  attractive  force  exerted  by  each  upon  the  other  has 
approximately  the  value 

yA7nxA7n.Jr2, 

r  being  the  distance  between  the  elements.  If  the  attraction  of  every 
element  of  mx  upon  every  element  of  m.l  be  computed  approximately 
in  the  same  manner,  the  resultant  of  these  forces  will  be  an  approx- 
imate value  of  the  resultant  attraction  of  mx  upon  m.x.  The  approx- 
imation is  closer  the  smaller  the  elements  into  which  the  bodies  are 
subdivided  ;  the  exact  value  is  the  limit  approached  by  the  approx- 
imate value  as  the  elements  approach  zero  in  size.  The  magnitude 
of  every  component  force  thus  approaches  zero  and  the  number  of 
components  approaches  infinity,  but  their  resultant  has  in  general  a 
finite  value.  The  computation  of  this  value  involves  a  process  of  in- 
tegration. 

In  applying  this  method  only  a  few  simple  cases  will  be  here  con- 
sidered. 

181.  Attraction  of  a  Homogeneous  Spherical  Shell  Upon  an 
Interior  Particle. — Proposition. —  The  resultant  attraction  exerted 
by  a  homogeneous  spherical  shell  of  uniform  thickness  upon  a  particle 
within  its  inner  surface  is  zero. 

Let  O  (Fig.  98)  be  the  position  of  the  particle,  and  through  O 


GRAVITATION.  1 59 

draw  any  straight  line  intersecting  the  outer  surface  of  the  shell  in 
two  points  A  and  B.  On  the  outer  surface  of  the  shell  take  an  ele- 
mentary area  containing  the  point  A  ;  from  every  point  of  the  per- 
imeter of  this  element  draw  a  line  through  0,  and  prolong  it  to  inter- 
sect the  spherical  surface  at  a  point  near  B.  We  thus  get  a  cone 
cutting  from  the  spherical  surface  at  B  a  second  elementary  area. 
Call  the  areas  of  the  elements  at  A  and  B,  a  and  b  respectively,  and 
let  h  be  the  thickness  of  the  shell.  The  volumes  cut  from  the  shell 
by  the  two  branches  of  the  conical  surface  are  approximately 

ha    and     hb 

respectively,  and  their  ratio  has  approximately  the  value 

ha/hb  =  a/b, 

the  approximation  being  closer  as  the  values  of  h,  a  and  b  are  taken 
smaller.      It  may  be  shown  that  as  a  and  b  approach  o, 

a/b   approaches    OA2/OB\ 

The  tangent  planes  to  the  sphere  at  A  and  B  are  equally  inclined 
to  the  line  AB ;  hence  in  the  limit,  as  a  and  b  approach  o,  they  are 
proportional  to  their  orthographic  pro- 
jections upon  a  plane  perpendicular  to 
AB  ;  that  is,  to  the  right  sections  at  A 
and  B  of  the  two  branches  of  the  con- 
ical volume  whose  vertex  is  at  O.  But 
these  right  sections  are  directly  propor- 
tional to  OA 2  and  0B%\  hence  the  areas 
a  and  b,  the  elementary  volumes  ha  and 
hb,  and  the  masses  of  these  elementary 
volumes, are  proportional  toOA2andOB'\ 

If  m  is  the  mass  of  the  particle  at  0, 
and  if  m'  and  m"  are  the  masses  of  the  elements  of  the  shell  at  A 
and  B  respectively,  the  particle  is  attracted  toward  A  with  a  force 

ymm'l  OA 2, 

and  toward  B  with  a  force 

ymm'/OB*. 

But,  as  just  shown, 

m'/mf=  OA2/OB\ 
or  m'/OA2=m"/OBi; 


i6o 


THEORETICAL    MECHANICS. 


hence  the  attractive  forces  of  m  and  m"  upon  m  are  equal  and  op- 
posite, and  their  resultant  is  zero. 

The  whole  volume  of  the  shell  may  be  divided  into  elements 
which,  taken  in  pairs,  are  related  in  the  same  way  as  the  two  ele- 
ments considered.  Hence  the  resultant  attraction  of  the  shell  upon 
the  particle  at  O  is  zero. 

In  the  above  reasoning,  it  was  assumed  that  the  thickness  of  the 
shell  becomes  small,  approaching  zero  as  a  limit.  If  the  thickness 
has  any  finite  value  the  result  still  holds.  For  the  shell  may  be 
regarded  as  made  up  of  a  great  number  of  very  thin  shells  for  each 
of  which  the  conclusion  is  true  at  least  approximately  ;  and  if  the 
number  of  shells  is  increased  without  limit,  the  thickness  of  each 
approaching  zero,  the  conclusion  holds  strictly  for  each  elementary 
shell  and  therefore  for  the  given  shell. 

The  proposition  is  also  true  if  the  density  of  the  shell  varies  in 
such  a  way  that  its  value  is  the  same  at  all  points  equally  distant  from 
the  center  of  the  sphere. 

182.  Attraction  of  a  Homogeneous  Shell  Upon  an  Exterior 
Particle. —  It  is  evident  from  symmetry  that  the  resultant  attraction 
of  a  homogeneous  spherical  shell  of  uniform  thickness  upon  an  ex- 
terior particle  is  di- 
rected along  the  line 
joining  the  particle 
with  the  center  of 
the  sphere. 

Let  A  (Fig.  99) 
be  any  point  of  the 
surface  of  the  shell, 
C  the  center  of  the 
sphere,  and  O  the 
position  of  the  ex- 
terior particle,  its  mass  being  m.  Take  an  element  of  the  spherical 
surface  containing  A,  and  let  a  denote  the  area  of  this  element,  h 
the  thickness  and  p  the  density  of  the  shell,  R  the  radius  of  the 
sphere.     Let    OA  =  r,    OC  =  c,   angle   A  CO  =  6,   angle   AOC 

=  </>• 

The  conical  surface  whose  base  is  a  and  apex  C  cuts  from  the 
shell  an  element  of  volume  whose  mass  is  pha.     The  attraction  of 


Fig.  99. 


GRAVITATION.  „  l6l 

this  mass  upon  the  particle  at  O  is  ymphajr2,  the  resolved  part  of 
which  in  the  direction  OC  is 

(ympAafr*)  cos  <f>. 

Let  the  element  of  area  a  be  taken  as  part  of  a  zone  of  infinitesimal 
width,  included  between  two  circles  whose  planes  are  perpendicular 
to  OC  For  every  point  of  such  a  zone,  r  and  <j>  have  the  same 
values  ;  hence  the  total  attraction  in  the  direction  OC  due  to  the 
portion  of  the  shell  corresponding  to  such  a  strip  is 

ympk  cos  cf>  dA/r2, 

if  dA  represents  the  area  of  the  zone.     The  attraction  due  to  the 
whole  shell  is  therefore 

cos  (j>  dA 


P  =  ymp/i  J 


the  integration   being  so  taken  as  to  cover  the  whole  surface  of  the 
sphere.     We  have  now  to  express  </>,  r  and  dA  in  terms  of  a  single 
variable  and  integrate.     Let  r  be  the  variable  chosen. 
We  have 

dA  =  Rdd  ■  2-jtR  sin  0  =  2-rrR2  sin  d  dd. 

But  r2  =  cl  +  Rl  —  2cR  cos  0 ;     rdr  =  cR  sin  0  d0  ; 

hence  dA  =  {2irRjc)r dr. 

Again,  R2  =  c2  +  r'1  —  2cr  cos  <f>, 

or  cos  </>  =  (V2  +  r2  —  R2)/2cr. 

Substituting  the  values  of  cos  <f>  and  dA  in  the  value  of  P,  and  ex- 
pressing the  proper  limits, 

_    irymp/iR  f*  +  Jt?  —  ^2  +  r2 


The  integral  expression  can  be  separated  into  two  parts,  thus : 


{c2  —  R2)  -  +  r. 
r 


ii 


1 62  THEORETICAL    MECHANICS. 

Taking  the  value  between  the  proper  limits, 

dr 


x 


'  +  *  c2  —  Rl  h 


c-  R  r 

-[(c-R)  -  (*+  R)]  +  [(r+  R)  -  (c-R)]  =  4R. 

Hence  P  =  TrymphR  •  4-R/c2  =  y  •  ^rrR2hp  •  m/c2. 

But  4.irR2hp  =  M  =  total  mass  of  shell.     Hence 

P=yMm/ci)  ....  (2) 
which  shows  that 

The  resultant  attraction  between  the  shell  and  an  exterior  par- 
ticle has  the  same  value  as  if  the  entire  mass  of  the  shell  were  con- 
centrated at  its  center. 

The  result  obviously  holds  for  a  shell  of  any  thickness  whose 
density  has  the  same  value  at  all  points  equally  distant  from  the 
center.  For  such  a  shell  may  be  subdivided  into  elementary  shells, 
each  of  which  may  be  taken  as  thin  and  as  nearly  of  uniform  density 
as  desired. 

Interior  particle.  —  The  reasoning  by  which  (1)  is  deduced  ap- 
plies also  to  the  case  of  an  interior  particle,  except  that  the  limits  of 
the  integration  must  be  different.  Thus,  for  the  attraction  on  an 
interior  particle,  we  have 

__  TrymphR  nRJcc2  —  Rl  +  r*      _ 
fc"       h-c  ~  r2 

which  agrees  with  Art.  181. 

183.  Attraction  of  a  Sphere  Upon  a  Particle. —  The  forego- 
ing results  may  be  applied  to  a  sphere  whose  density  has  the  same 
value  at  all  points  equally  distant  from  the  center. 

Exterior  particle. —  The  resultant  attraction  of  such  a  sphere 
upon  an  exterior  particle  has  the  same  value  as  if  the  entire  mass  of 
the  sphere  were  concentrated  at  its  center. 

Interior  particle. —  The  resultant  attraction  of  such  a  sphere  upon 
an  interior  particle  is  the  same  as  if  the  portion  of  the  mass  nearer 
the  center  than  the  particle  were  concentrated  at  the  center  ;  the  re- 
maining shell  being  disregarded  because  its  resultant  attraction  is 
zero. 


gravitation.  1 63 

Examples. 

1 .  Assuming  the  earth  to  be  a  sphere  whose  density  is  a  function 
of  the  distance  from  the  center,  compare  the  weight  of  a  body  at  the 
surface  and  at  a  point  h  feet  above  the  surface. 

The  ' '  weight "  of  a  body  is  the  force  with  which  the  earth  at- 
tracts it.*  If  R  is  the  radius  of  the  earth,  and  if  Pand  P*  are  the 
values  of  the  earth's  attraction  upon  the  body  at  the  surface  and  at 
h  feet  above  the  surface  respectively,  we  have 

P/P*  =  (R  +  k)*IR\ 
or  P'  =  PR'2/(R  +  h)\ 

Hence  the  weight  of  a  body  at  height  h  above  the  surface  is  the 
fraction  R2/(R  -\-  h)2  of  its  weight  at  the  surface. 

If  h  is  a  small  fraction  of  R,  we  have,  approximately, 
R%R  +  h)2  =  (1  +  k/R)-2  =  1  —  2/1 1 R. 

2.  At  what  height  above  the  surface  will  the  weight  of  a  body  be 
1  per  cent  less  than  at  the  surface  ?  [The  mean  radius  of  the  earth 
is  very  nearly  6.3709  X  io8  cm.  or  20,902,000  ft.] 

3.  If  the  pound-force  is  defined  as  the  weight  of  a  pound-mass 
at  the  earth's  surface,  how  much  does  a  change  of  elevation  of  10,000 
ft.  affect  the  value  of  this  unit  force  ? 

4.  Let  G  denote  the  weight  of  a  given  mass  at  the  earth's  sur- 
face, and  G'  its  weight  at  a  depth  h  below  the  surface.  If  the  earth 
were  a  sphere  of  uniform  density  and  of  radius  R,  show  that 

G'/G  =  (R  —  h)IR  =  1  —  k/R. 

5.  Assuming  that  the  earth  is  a  sphere  and  that  the  density  is  a 
function  of  the  distance  from  the  center,  let  p  denote  the  mean  den- 
sity of  the  whole  earth  and  p0  the  mean  density  of  the  outer  shell  of 
thickness  h.  Determine  the  relation  between  the  weight  of  a  body 
at  the  surface  and  its  weight  at  a  depth  h  below  the  surface. 

Let  M  be  the  mass  of  the  whole  earth,  M '  that  of  the  inner  sphere 
of  radius  R  —  k,  m  the  mass  of  the  given  body,  G  its  weight  at 
the  surface  and  G'  its  weight  when  at  depth  k  below  the  surface. 
Then  G  is  equal  to  the  attraction  between  two  particles  of  masses  M 
and  in  whose  distance  apart  is  R  ;  and  G'  is  equal  to  the  attraction 
between  two  particles  of  masses  M'  and  ;;/  whose  distance  apart  is 
R  —  h.     That  is,  G  =  yMm/R2,  G'  =  yM'mfcR  —  h)\     Hence 

£=*1-*1Y\     .     .     .  (o 

G       M\R  —  h) 
But       M  =  4ttR*p  ;     M—M'  =  47rp0 [R*  —  (R  —  hf\ 


*  The  effect  of  the  earth's  rotation  upon  the  apparent  weight  is  here  dis- 
regarded.   See  Art.  311. 


164  THEORETICAL    MECHANICS. 


M   ,  pjR 


C-(^)H-?K6( 


M'=      _gj\      lR  —  h\* 

M    '  *  "    p 
Substituting  this  value  in  equation  (1) 

G'        I  p»\(      R      V       Po(R 


R 


(2) 


H'-3fe^r+?m  •  ■ * 


Approximate  simple  formula. —  \i  h  is  a  small  fraction  of  R, 
equation  (3)  may  be  reduced  to  the  following  as  a  first  approxima- 
tion : 

^=i+(2-w  •   •   • (4) 

6.  Show  that,  if  the  mean  density  of  the  outer  layer  of  the  earth 
is  less  than  two-thirds  the  mean  density  of  the  whole  earth,  the 
weight  of  a  body  increases  as  it  is  carried  below  the  surface. 

This  follows  from  equation  (4)  above.  It  will  be  seen  that,  as  a 
body  is  carried  below  the  surface  of  the  earth,  its  change  of  weight  is 
the  resultant  of  two  opposite  effects.  The  total  mass  attracting  it  is 
less  because  the  outer  shell  has  on  the  whole  no  effect  (Art.  181)  ; 
while  the  attraction  of  the  inner  sphere  is  greater  because  the  body  is 
nearer  its  center  (Art.  183).  The  latter  effect  is  greater  than  the 
former  unless  the  mean  density  of  the  outer  shell  is  at  least  two- 
thirds  that  of  the  whole  earth. 

According  to  the  best  determinations  the  mean  density  of  the 
earth  is  very  nearly  5.527  times  that  of  water,  while  the  average  den- 
sity of  rocks  near  the  surface  may  be  taken  at  about  half  this  value. 
If  p0/p  =  1/2,  equation  (4)  becomes 

G'/G^i  +k/2R.  .        .        .     (5) 

That  the  weight  of  a  body  increases  when  it  is  carried  below  the 
surface  of  the  earth  has  been  shown  by  experiment. 

7.  What  would  be  the  change  in  the  weight  of  a  body  if  carried 
2,000  ft.  below  the  earth's  surface,  assuming  the  density  of  the  outer 
portion  of  the  earth  to  be  half  the  mean  density  ? 

184.  Value  of  the  Constant  of  Gravitation. — If  the  attrac- 
tion between  two  particles  of  known  mass  at  a  known  distance  apart 
can  be  determined  by  experiment,  the  value  of  7,  the  constant  of 
gravitation,  can  be  determined.  From  the  foregoing  results  (Arts. 
182,  183),  spheres  of  any  size  may  be  used  instead  of  particles.  The 
attraction  between  bodies  of  ordinary  size  is  so  small  that  it  can  be 
measured  only  by  the  most  delicate  apparatus. 

The  relation  between  the  constant  of  gravitation  and  the  mass  of 
the  earth  may  be  shown  as  follows,  assuming  the  earth  to  be  a  sphere 
whose  density  is  a  function  of  the  distance  from  the  center,    We  shall 


GRAVITATION.  1 65 

take  as  units  those  of  the  British  gravitation  system.  The  unit  mass 
is  the  pound,  the  unit  force  the  weight  of  a  pound  mass  at  the  earth's 
surface  (the  pound-force),  the  unit  length  the  foot. 

Let  R  denote  the  radius  of  the  earth  in  feet,  M  its  mass,  p  its 
mean  density.  Consider  the  attraction  of  the  earth  upon  a  body  of 
mass  m  at  the  surface. 

By  the  general  formula  of  Art.  176  the  value  of  this  attraction  is 

yMm/R2.      But  expressed   in  pounds-force  it  is  also  equal  to  m. 

Hence  m  =  yMm/R2,  or 

yM  =:  R\ 

Since  the  value  of  R  is  known,  this  equation  serves  to  determine 
either  of  the  quantities  7  and  M  when  the  other  is  known. 

Instead  of  M,  we  may  introduce  the  earth's  mean  density  p,  since 
M  =  ^ttR^p/t,'     The  equation  then  becomes 

vp  =  sIattR. 

The  best  determinations  give  as  the  earth's  mean  density  about 
345  pounds  per  cubic  foot.  Taking  R  =  20,900,000  ft.,  the  value 
of  the  constant  of  gravitation  *  is 

7  =  3/47^/0  =  3/(47T  X  20,900,000  X  345)  =  3-31  X  io-u. 
185.  Value  of  Gravitation  Unit  of  Mass. —  Let  m  pounds  be 
the  mass  of  each  of  two  particles  which,  placed  one  foot  apart, 
attract  each  other  with  one  pound-force.  The  value  of  m  may  be 
found  from  the  formula  P  =  ymxmjr2}  by  putting  P  =  1,  r  =  1, 
mx  =  m%  =  m,  and  7  =  3.31  X  io~u  as  found  above.     The  result 

1S  m  =  i/j/7  =  173,800  pounds. 

If  a  mass  equal  to  173,800  pounds  be  taken  as  the  unit  mass,  the 
constant  7  becomes  unity,  and  the  formula  for  the  attraction  between 
two  particles  whose  masses  are  mx  and  m2  and  whose  distance  apart 

isris  P=mxmjr\ 

Thus,  the  "  gravitation  unit  of  mass"  (Art.  178)  is  a  mass  equal  to 
about  173,800  pounds,  if  distance  is  expressed  in  feet  and  force  in 
pounds-force. 

*The  above  is  not  given  as  the  actual  method  of  determining  T,  but 
merely  as  showing  about  what  its  value  is.  The  above  value  of  the  earth's 
mean  density  is  based  upon  the  experimental  determination  of  T  by  the  meas- 
urement of  the  attraction  between  bodies  of  known  mass. 


1 66  THEORETICAL    MECHANICS. 

1 86.  Values  of  the  Constant  of  Gravitation  and  of  the  Grav- 
itation Unit  of  Mass  in  the  C.  G.  S.  System.* — If  the  unit  mass 
is  the  gram,  the  unit  length  the  centimeter  and  the  unit  force  the 
dyne  (Art.  217),  the  value  of  7  may  be  found  as  follows  : 

The  attraction  of  the  earth  for  a  body  of  m  grams  mass  at  the 
surface  is  approximately  98 \m  dynes.  By  the  law  of  gravitation  it 
is  also  equal  to  yMm/R2.     Hence  981^  =  jMm/R2,  or 

yM=  9S1R2. 

Here  M  is  the  mass  of  the  earth  in  grams,  and  R  is  its  radius  in 
centimeters.  Introducing  the  earth's  mean  density  p  instead  of  M, 
the  equation  is 

7P  =  3X  981/47^. 

Taking  p  =  5.527  grams  per  cubic  centimeter  and  R  =  6.371 
X  1  o8  centimeters, 


—8 


7  =  (3  X  98i)/(4T  X  5-527  X  6.371  X  io8)  =  6.65  X  10 

Let  m  grams  be  the  mass  of  each  of  two  particles  which,  when  1 
centimeter  apart,  attract  each  other  with  a  force  of  1  dyne.     Then 

1  =  <ym2, 


or  m  =  1/1/7  =  1/^6.65  X  10^=  3,880  grams. 

If  a  mass  of  3, 880  grams  is  taken  as  the  unit  mass,  and  if  the  centi- 
meter is  the  unit  length,  the  attraction  in  dynes  between  two  particles 
whose  masses  are  mx  and  m2  and  whose  distance  apart  is  r  is  given 
by  the  formula 

P  =  m^mjr2. 

The  above  relation  between  the  constant  of  gravitation  and  the 
earth's  mean  density  is  only  approximate,  since  the  earth  does  not 
attract  bodies  at  the  surface  exactly  as  if  its  mass  were  concentrated 
at  its  center.  The  value  of  the  earth's  mean  density  above  given 
is  that  of  Professor  C.  V.  Boys.f  It  is  based  upon  the  value 
7  =  6.6576  X  io"8. 

*  The  "absolute"  or  kinetic  system  of  units  is  explained  in  Chapter  XII. 
The  present  Article  presupposes  a  knowledge  of  this  system. 

fSee  "Nature,"  Vol.  L,  p.  419.  For  values  found  by  other  experi- 
menters see  "Nature,"  Vol.  LV,  p.  296. 


Part  II. 
MOTION   OF   A   PARTICLE. 


CHAPTER   XII. 

MOTION   IN   A   STRAIGHT    LINE  :     FUNDAMENTAL   PRINCIPLES. 

§  i.   Position,  Displacement  and  Velocity. 

187.  Position  of  a  Particle  in  a  Given  Line. —  If  a  particle 
moves  in  a  given  straight  line,  its  position  at  any  instant  is  known  if 
its  distance  from  a  given  fixed  point  in 

the  line  is  known.     Let  0  (Fig.  100)     ■*------  x    -», 

be  the  fixed  point,  and  P  the  position     P'  0  P 

of  the  particle  at  any  instant.      Let  the  Fig.  100. 

distance  OP  be  represented  by  x  ;  then 

the  position  of  the  particle  is  specified    by  the  value  of  x.      The 
quantity  x  is  called  the  abscissa  of  the  particle. 

The  value  of  x  involves  direction  as  well  as  distance.  The  two 
directions  OP,  OP'  along  the  line  of  motion  are  distinguished  as  plus 
and  minus. 

188.  Displacement. —  The  change  of  position  of  a  particle  dur- 
ing any  interval  of  time  is  called  its  displacement.  If  A  and  B  (Fig. 
101)  are  its  positions  at  the  beginning  and  end  of  the  interval,  AB 

is  its  displacement.     The  displacement, 

*.  x     *    l  like  the  abscissa,  is  called  plus  or  minus 

^ -4  -g        according  to  its  direction  along  the  line 

Fig.  .01.  of  motion- 

If  xx  and  xt  are  the  values  of  x  at  the 

beginning  and  end  of  the  interval  respectively,  x2  —  xx  gives  the 

displacement  in  magnitude  and  sign. 

189.  Uniform  Motion. —  The  motion  of  a  particle  is  uniform  if 
it  receives  equal  displacements  in  any  equal  intervals  of  time,  however 
small  these  intervals  may  be  taken. 

190.  Velocity. —  The  velocity  of  a  particle  is  its  rate  of  moving 
(or  rate  of  displacement). 


1 68  THEORETICAL    MECHANICS. 

If  the  particle  moves  uniformly  throughout  a  certain  interval,  its 
velocity  is  proportional  directly  to  its  total  displacement  during  the 
interval  and  inversely  to  the  duration  of  the  interval. 

The  numerical  value  of  the  velocity  depends  upon  the  unit  in 
terms  of  which  it  is  expressed. 

191.  Unit  Velocity. —  Any  "rate  of  moving"  might  be  chosen 
as  the  unit  velocity,  but  it  is  convenient  to  choose  as  the  unit  the 
velocity  possessed  by  a  uniformly  moving  particle  which  passes  over 
the  unit  distance  in  the  unit  time. 

The  unit  thus  defined  is  a  derived  unit  (Art.  13)  whose  value 
depends  upon  the  units  of  distance  and  time. 

With  the  foot  and  second  as  the  fundamental  units,  the  unit  ve- 
locity is  the  foot-per-second.  This  is  the  unit  ordinarily  employed  in 
engineering  applications  in  those  countries  in  which  the  British  sys- 
tem is  in  common  use. 

When  the  French  metric  system  is  used,  the  meter  takes  the 
place  of  the  foot  as  the  engineers'  unit  of  length,  and  the  correspond- 
ing unit  of  velocity  is  the  meter-per -second. 

In  purely  scientific  work,  the  centimeter  is  almost  universally 
adopted  as  the  unit  length,  the  corresponding  unit  velocity  being  the 
centimeter-per-second.     (See  Art.  217.) 

Dimensions  of  unit  velocity. — Since  the  unit  velocity  is  so  de- 
fined as  to  be  proportional  directly  to  the  unit  length  and  inversely 
to  the  unit  time,  there  may  be  written  the  dimensional  equation 

V  =  L/T, 

if  V  denotes  the  unit  velocity,  L  the  unit  length  and  T  the  unit  time 
(Art.  15). 

192.  Numerical  Value  of  Velocity. —  The  numerical  value  of 
the  velocity  of  a  particle  which  moves  uniformly  throughout  a  certain 
interval,  expressed  in  terms  of  the  unit  above  defined,  is  found  by 
dividing  the  length  of  the  displacement  by  the  duration  of  the  in- 
terval. 

Thus,  let  x  denote  the  abscissa  of  a  uniformly  moving  particle  at 
the  time  /,  reckoned  from  some  assumed  instant  (taken  as  origin  of 
time).  Let  xx  and  xt  be  the  values  of  x  at  times  tx  and  /.,.  Then  if  v 
denotes  the  velocity, 

V  =  (x2  —  xj/it,  —  tj. 


MOTION    IN    A    STRAIGHT    LINE.  1 69 

If  x  and  /  are  expressed  in  feet  and  seconds  respectively,  the 
value  of  v  given  by  this  formula  is  in  feet-per-second.  If  x  is  in  cen- 
timeters, v  is  in  centimeters-per-second. 

I93»  Sign  of  Velocity.  —The  velocity  will  be  called  plus  or  minus 
according  as  the  particle  moves  in  the  positive  or  in  the  negative 
direction  along  the  line  of  motion. 

The  formula  v  =  (x2  —  #i)/(t.2  —  A  )  gives  the  value  of  the  ve- 
locity in  sign  as  well  as  in  magnitude  ;  for  x2  —  xx  will  be  positive  if 
the  motion  has  the  plus  direction  and  negative  if  the  motion  has  the 
minus  direction. 

194.  Variable  Motion. —  If  a  particle  receives  unequal  displace- 
ments in  equal  intervals  of  time,  its  motion  is  variable.  Its  velocity 
may  still  be  defined  as  its  rate  of  moving,  but  the  above  method  of 
computing  the  value  of  this  rate  becomes  inapplicable. 

Velocity  must  now  be  understood  to  be  a  quantity  which  has  a 
definite  value  at  any  instant,  but  whose  value  continually  varies. 
The  meaning  of  velocity  in  case  of  variable  motion  is  best  explained 
by  a  consideration  of  ' '  average  velocity. ' ' 

195.  Average  Velocity. —  If  a  particle  moves  in  any  manner,  its 
average  velocity  during  any  given  interval  may  be  defined  as  the 
velocity  of  a  particle  which,  moving  uniformly,  receives  an  equal  dis- 
placement in  the  same  interval. 

The  formula 

v  =  (x2—x1)/(t2-tl) 

always  gives  the  value  of  the  average  velocity  for  the  interval  from 


196.  Approximate  Value  of  Velocity  at  an  Instant. —  The 
velocity  of  a  particle  at  any  instant  may  be  computed  approximately 
by  finding  the  average  velocity  for  a  very  short  interval. 

Let  x  denote  the  abscissa  of  the  particle  at  the  instant  t,  and  let 
Ax  denote  the  increment  of  x  in  a  short  interval  of  time  At ;  that  is, 
Ax  is  the  displacement  during  the  time  At.  Then  Ax/ At  is  an 
approximate  value  of  the  velocity  at  the  instant  t.  The  approxima- 
tion is  closer  the  shorter  the  interval  At. 

197.  True  Value  of  the  Velocity  at  an  Instant. —  The  true 
value  of  the  velocity  at  the  time  /  is  the  limit  approached  by  the 


I70  THEORETICAL   MECHANICS. 

approximate  value  as  the  assumed  interval  is  taken  smaller,  approach- 
ing zero.     But  this  limit  is  the  derivative  of  x  with  respect  to  t,  that  is-, 

v  =  lim  [Ax/Af]  =  dxjdt. 

This  formula  is  the  mathematical  expression  of  the  definition  of  ve- 
locity at  an  instant. 

198.  Computation  of  Velocity  When  Variable. —  If  the  position 
of  a  particle  is  known  at  every  instant,  the  velocity  at  any  instant 
may  be  computed  from  the  above  formula.  Thus,  if  x  is  known  as 
a  function  of  t,  dxjdt  may  be  determined  by  differentiation. 

Examples. 

1.  The  position  of  a  particle  is  given  by  the  equation  x  =  &t'\  k 
being  a  constant. 

■*  (a)  Show  that  the  velocity  at  any  instant  is  given  by  the  formula 

V  =  2kt. 

(b~)  Where  is  the  particle  at  the  instant  taken  as  "origin  of 
time"? 

(V)  If,  3  sec.  after  the  ' '  origin  of  time, ' '  the  particle  is  1 2  ft. 
from  the  origin  from  which  x  is  measured,  what  is  the  value  of  the 
constant  k  ? 

(d)  Assuming  condition  (V),  what  is  the  velocity  when  the  par- 
ticle is  10  ft.  from  the  origin? 

(V)  What  is  the  value  of  k  if  the  centimeter  is  taken  as  the  unit 
length  ? 

(/)  What  is  the  velocity  when  the  particle  is  1  met.  from  the 
origin?     (Express  the  result  in  centimeter-second  units.) 

2.  The  position  of  a  particle  is  given  by  the  equation  x  =  2/  -j-  t'\ 
x  being  in  feet  and  t  in  seconds. 

{a)  Compute  the  average  velocity  for  each  of  the  following  inter- 
vals, beginning  at  the  instant  t  =  10:  2  sec,  1  sec,  0.5  sec,  o.  1 
sec,  0.0 1  sec,  0.001  sec. 

(J?)  Compute  the  exact  value  of  the  velocity  at  the  instant  t  =  10. 

Ans.  {a)  The  values  of  the  average  velocity  in  ft.  -per-sec  are  24, 
23,  22.5,  22.1,  22.01,  22.001.     {b)  22  ft. -per-sec 

3.  The  position  of  a  body  falling  freely  from  rest  is  given  approx- 
imately by  the  equation  x  =  16.  it2,  in  which  x  ft.  is  the  distance 
fallen  in  /  sec.  Compare  the  true  velocity  at  the  end  of  2  sec.  with 
the  average  velocity  for  an  interval  of  o.  1  sec.  after  the  instant  t  =  2 . 

Ans.  True  vel.  =  64.4  ft. -per-sec  Average  vel,  =  66.01  ft- 
per-sec 

199.  Representation  of  Velocities  by  Lines. —  If  P  (Fig.  102) 
represents  the  position  of  a  particle  at  any  time  /  and  O  is  the  as- 


MOTION    IN    A   STRAIGHT    LINE. 


I7I 


sumed  origin  in  the  line  of  motion,  OP  represents  x,  the  abscissa 
of  the  particle  at  the  time  /.  On  another  line  parallel  to  OP,  take 
an  origin  O',  and  lay  off  O'P'  equal  (on    • 

any  convenient  scale)  to  v,  the  velocity  of         K x     „ 

the  particle  at  the  time  t.     If  the  velocity         q  p 

is  constant,  the  point  P'  remains  fixed  in  „_..    v  =  ±    ...  * 

position ;  if  the  velocity  of  P  varies,  P'  q>  p~' 

moves.      If  the    motion   of  P'   is   known,  Fig.  102. 

the   velocity   at   every   instant   is   known. 

200.  Notation  for  Time-Derivatives. —  When  a  variable  quan- 
tity is  a  function  of  the  time,  its  derivatives  are  often  designated  by 
the  use  of  dots,  as  follows  : 

dx/dt  =  x ;     d^x/dt*  =  dx/dt  =  x ;     etc. 
This  notation  will  often  be  used  in  the  following  pages. 

§  2.    Velocity-Increment  and  Acceleration. 

201.  Increment  of  Velocity. —  Let  vx  and  v2  be  the  values  of 
the  velocity  at  instants  tx  and  t2  respectively ;  then  v2  —  vx  is  the 
velocity-increment  for  the  interval  from  tx  to  t.2 .  This  increment  may 
be  either  positive  or  negative,  according  as  vt  is  (algebraically) 
greater  or  less  than  vx . 

Geometrical  representation. — If,  as  described  in  Art.  199,  the 
velocity  at  every  instant  be  represented  in  magnitude  and  direction 
by  a  length  O'P'  laid  off  from  a  fixed  point  O'  (Fig.  102),  the  point 
P'  will  move  as  the  velocity  varies.  When  v  increases  (algebraic- 
ally), P'  moves  in  the  plus  direction  ;  when  v  decreases,  P'  moves  in 
the  minus  direction.  The  displacement  of  P'  in  any  interval  repre- 
sents the  velocity-increment. 

The  velocity -increment  must  not  be  confounded  with  the  displace- 
ment of  the  particle.  There  is  not  necessarily  any  relation  between 
them.  The  point  P'  may  or  may  not  move  in  the  same  direction  as 
the  particle. 

202.  Uniformly  Varying  Velocity. —  If  the  velocity  receives 
equal  increments  in  all  equal  intervals  of  time,  however  small  these 
may  be,  the  velocity  is  uniformly  variable. 

Recurring  to  the  geometrical  representation  given  above  (Fig. 
102),  if  the  velocity  of  P  changes  at  a  uniform  rate,  the  point  P 


172  THEORETICAL    MECHANICS. 

moves  uniformly.  The  motion  of  P'  has  the  plus  or  the  minus 
direction  according  as  the  velocity  of  P  is  (algebraically)  increasing 
or  decreasing. 

203.  Acceleration. —  The  rate  of  change  of  the  velocity  of  a  par- 
ticle is  called  its  acceleration. 

If,  throughout  a  certain  interval  of  time,  the  velocity  changes  at  a 
uniform  rate,  the  acceleration  is  proportional  directly  to  the  total 
velocity-increment  and  inversely  to  the  duration  of  the  interval. 

The  numerical  value  of  the  acceleration  depends  upon  the  unit  in 
which  it  is  expressed. 

204.  Unit  of  Acceleration. — Although  the  unit  of  acceleration 
may  be  chosen  arbitrarily,  it  is  convenient  to  make  it  depend  upon 
the  units  of  length  and  time.  Hence  the  unit  acceleration  is  defined 
as  that  of  a  particle  whose  velocity  increases  by  one  unit  in  every 
unit  of  time. 

Thus,  with  the  foot  and  second  as  fundamental  units,  the  unit 
acceleration  is  that  of  a  particle  whose  velocity  increases  by  one  foot- 
per-second  in  each  second. 

With  the  meter  replacing  the  foot  as  the  unit  length,  the  unit 
acceleration  is  that  of  a  particle  whose  velocity  increases  by  one 
meter-per-second  in  each  second. 

Similarly,  a  third  unit  of  acceleration  is  derived  from  the  centi- 
meter and  second  as  fundamental  units. 

Dimensions  of  unit  acceleration. — The  above  definition  of  the 
unit  acceleration  makes  it  proportional  directly  to  the  unit  velocity 
and  inversely  to  the  unit  time.     Representing  the  unit  acceleration 
by  A,  there  may  be  written  the  dimensional  equation 
A  =  V/T  =  L/T2. 

205.  Numerical  Value  of  Acceleration. — When  the  velocity  of 
a  particle  increases  at  a  uniform  rate  during  any  interval,  the  value  of 
the  acceleration  in  terms  of  the  unit  above  denned  is  computed  by 
dividing  the  increment  of  velocity  by  the  duration  of  the  interval. 
Thus,  let  v  denote  the  velocity  at  any  instant  /,  vx  and  v2  the  values 
of  v  at  the  instants  tx  and  tt ,  and  p  the  acceleration.     Then 

P=(v2  —  vl)/(t.2-tl). 
This  value  ofp  may  be  in  feet-per -second- per-second,  in  meters- 
per-second-per-second,  in   centimeters-per-second-per-second,  or  in 
other  units,  according  to  the  units  in  which  t  and  v  are  expressed. 


MOTION    IN    A   STRAIGHT    LINE.  173 

206.  Algebraic  Sign  of  Acceleration. — If  v2  —  vx  is  negative, 
the  value  of  p  is  negative.  This  means  that  the  velocity  is  decreasing 
algebraically  ;  that  is,  that  the  velocity-increment  received  by  the 
particle  in  any  interval  has  the  minus  direction.  If  p  is  plus,  the 
velocity  is  increasing  algebraically,  although  numerically  it  may  be 
decreasing.     Thus, 

if  v  is  positive   and  increasing  in  magnitude,  p  is  positive  ; 
"  V  ' '  negative     ■  '    decreasing  ' '  "  fM  •.*.-.• 

\f-v ■"  positive      "   decreasing  M  M  /r"  negative  ; 

1 '  v  "  negative     ' '    increasing   "  "         p  "  ' ■    . 

207.  Non-Uniform  Variation  of  Velocity. —  In  case  the  incre- 
ments of  velocity  in  equal  intervals  of  time  are  unequal,  acceleration 
may  still  be  defined  as  the  rate  of  change  of  the  velocity,  but  the 
above  method  of  computing  its  value  becomes  inapplicable. 

Acceleration  must  now  be  understood  to  be  a  variable  quantity, 
having  a  definite  value  at  any  instant.  Its  meaning  is  best  under- 
stood by  a  consideration  of  ' '  average  acceleration. ' ' 

208.  Average  Acceleration. —  The  average  acceleration  of  a  par- 
ticle for  an  interval  during  which  its  velocity  varies  in  any  way  is  an 
acceleration  which,  if  constant,  would  result  in  the  same  velocity- 
increment  in  the  same  interval. 

The  formula 

/  as  (v,  —  v,)i{t2  -  O 

always  gives  the  value  of  the  average  acceleration  for  the  interval 
from  tx  to  tt. 

209.  Approximate  Value  of  Acceleration  at  an  Instant. — An 
approximate  value  of  the  acceleration  at  an  instant  may  be  found  by 
computing  the  average  acceleration  for  a  very  short  interval  of  time. 

Let  v  denote  the  velocity  at  an  instant  /,  and  &v  the  velocity- 
increment  for  a  short  time  At.  Then  Av/At  is  an  approximate 
value  of  the  acceleration  at  the  instant  /.  The  approximation  is  closer 
the  shorter  the  interval  At. 

210.  Exact  Value  of  Acceleration  at  an  Instant. —  The  exact 
value  of  the  acceleration  at  the  time  /  is  the  limit  approached  by  the 
approximate  value  as  At  is  taken  smaller  and  smaller,  approaching 
zero.     This  limit  is  the  derivative  of  v  with  respect  to  / ;  that  is, 

/  =3  lim  \Av\kti\  =s  dv/dt. 


174  THEORETICAL    MECHANICS. 

Since  v  =  dx/dt  =  x,  we  have 

p  =  (Px\dt%  —  dx/dt  =  jr. 

211.  Application  of  Formulas  for  Velocity  and  Accelera- 
tion.— The  formulas 

V  =  dx/dt,    p  =  dvjdt  *=  d2xjdt\ 

are  of  fundamental  importance  in  the  solution  of  various  problems  in 
rectilinear  motion.  Suppose,  for  example,  that  x  is  known  as  a  func- 
tion of  / ;  that  is,  that  the  position  at  every  instant  is  known.  In 
this  case  v  and  p  can  be  determined  by  differentiation.  Thus,  if 
x  =  f(f),  we  have 

v  =  dxldt  =  df{t)jdt  =f'(f), 

p  ==  dvjdt  =  d*xjdtl  =  df\t)jdt  =/"(*•}. 

As  another  case,  suppose  the  velocity  or  acceleration  known  at 
every  instant ;  that  is,  let  v  or  p  be  a  known  function  of  /.  In  this 
case  the  value  of  x  in  terms  of  /  can  be  found  by  integration. 

In  general,  if  a  relation  is  known  between  any  or  all  of  the  quan- 
tities x,  t,  v  and  /,  the  motion  can  be  completely  determined  either 
by  differentiation  or  by  integration.  If  the  solution  involves  integra- 
tion, either  one  or  two  arbitrary  constants  will  be  brought  in,  the 
values  of  which  cannot  be  determined  without  additional  data  as  to 
the  conditions  of  the  motion. 

Thus,  if  there  is  given  such  a  relation  as 

fix,  /,  v,  p)  ==  o, 

the  substitution  of  the  general  values  of  v  and  p  as  derivatives  with 
respect  to  t  gives  a  differential  equation  whose  solution  determines 
the  motion. 

Examples. 

i.  The  position  of  a  particle  is  given  by  the  equation  x  =  kt'\  k 
being  a  constant.  Show  that  the  acceleration  is  constant,  and  deter- 
mine its  value  in  terms  of  k. 

2.  In  Ex.  i,  if  the  velocity  is  8  ft.-per-sec.  at  the  instant  t  =  3, 
what  are  the  values  of  k  and  of  the  acceleration  ? 

3.  If  the  position  of  a  particle  is  given  by  the  equation  x  =  kt2 
+  k't  +  k" ',  show  that  the  acceleration  has  the  same  value  as  if  the 
equation  were  x  =  kt'1. 

4.  The  position  of  a  particle  is  given  by  the  equation  x  =  10  -J-  5^ 
+  2/2  -\-  t'\  x  being  in  feet  and  /  in  seconds.    Compute  the  average 


MOTION    IN    A   STRAIGHT    LINE. 


175 


acceleration  for  o.  1  sec.  after  the  instant  /  ==  2,  and  compare  it  with 
the  exact  value  of  the  acceleration  at  the  beginning  of  the  interval. 
The  velocity  is  given  by  the  formula 

v  =  x  =  5  +  4/  -j-  3/2. 

When  /  =  2,  v  =  25.  When  /  =  2. 1,  V  -==  26.63.  Hence  the  av- 
erage acceleration  is 

Av/At  =  (26.63  —  25)/°-1  ==  x6-3  ft. -per-sec. -per-sec. 

The  true  acceleration  at  any  instant  is 

/  =  dvjdt  =  x  =  4.  -\-  6t. 

When  /  =  2,  p  =  16  ft. -per-sec. -per-sec. 

5.  If  the  velocity  of  a  particle  is  given  by  the  equation  v  =  12  — 
16/,  determine  the  position  at  any  time. 

The  equation  may  be  written 

i:  =  dxjdt  =12  —  16/. 

Integrating  with  respect  to  /, 

X  =  12t  —  8t2  -f  C. 

In  order  that  the  constant  of  integration  C  may  be  determined,  the 
position  of  the  particle  at  some  given  instant  must  be  known.  Thus, 
if  t  is  reckoned  from  the  instant  at  which  the  particle  is  at  the  origin, 
the  values  (x  =  o,  /  =  o)  must  satisfy  the  equation.  This  requires 
that  C==  o.  But  if  the  " origin  of  time"  (z.  e.,  the  instant  from 
which  t  is  reckoned)  is  chosen  as  the  instant  when  x  =  4,  the  values 
(x  =  4,  t  =  o)  must  satisfy  the  equation,  and  therefore  C  =  4.  If 
jr  is  in  feet,  in  what  units  is  C? 

6.  If  v=  12  —  16/,  determine  the  acceleration  at  the  instant 
t  =  5,  and  the  average  acceleration  for  o.  1  sec.  after  t  =  5. 

7.  Let  the  velocity  be  given  by  the  equation  v  =  12  -f-  18/2,  / 
being  in  seconds  and  7/  in  centimeters-per-second.  Determine  the 
position,  the  velocity  and  the  acceleration  at  the  instant  t  =10  sec. 
In  order  that  the  value  of  x  may  be  completely  determined,  what 
additional  data  must  be  given  ? 

8.  With  data  of  Ex.  7,  determine  the  average  velocity  and  the 
average  acceleration  during  o.  5  sec.  and  during  o.  1  sec.  following  the 
instant  /  =  10  sec. 

9.  The  velocity  of  a  particle  at  a  certain  instant  is  too  ft. -per-sec. 
in  the  positive  direction.  The  acceleration  is  constant  and  equal  to 
24  ft. -per-sec. -per-sec.  in  the  negative  direction,  {a)  When  will  the 
particle  be  at  rest  ?     {B)  What  will  be  its  velocity  4  sec.  later  ? 

Ans.  (a)  At  the  end  of  4^  sec.     (<£)  — 96  ft. -per-sec. 

10.  The  acceleration  being  constant,  determine  the  position  and 
velocity  at  any  time. 


176  THEORETICAL    MECHANICS. 

Let  f  be  the  given  constant  value  of  p.  Then  the  problem  is  to 
solve  the  differential  equation 

d%x\dt%  =■  dvjdt  =  f. 

Integrating  with  respect  to  t, 

v  =  dx/dt  =  ft  +  C, 

C  being  a  constant  of  integration.  To  determine  its  value,  an  ' '  initial 
condition  "  must  be  known  ;  that  is,  the  value  of  the  velocity  at  some 
definite  instant  must  be  known.  Let  v0  denote  the  value  of  v  when 
/  =  o  ;  then,  since  the  last  equation  is  true  for  any  simultaneous 
values  of  v  and  /, 

vo  —  /•  °  +  C ;     or     C  =  v0. 

Hence  v  =  x  =ft  -\-  v0. 

Integrating  the  last  equation  with  respect  to  /, 

x  =  \fi%  +  v  +  c. 

To  determine  the  constant  of  integration,  C\  the  position  at  some 
given  instant  must  be  known.  Let  x0  denote  the  value  of  x  when 
t  =  o ;  then,  the  last  equation  being  true  for  x  =  x0  and  /  =  o, 
C  must  equal  x0 .     Hence 

x=\ftl  -fV+iJ. 

11.  The  velocity  of  a  falling  body  is  observed  to  increase  by  32.2 
ft.-per-sec.  during  every  second  of  its  motion.  How  far  will  it  fall 
from  its  position  of  rest  in  t  seconds  ? 

[The  acceleration  is  constant  and  equal  to  32.2  ft.-per-sec.-per- 
sec.     Hence  this  is  a  special  case  of  Ex.  10.] 

12.  The  acceleration  of  a  particle  increases  in  direct  ratio  with 
the  time  reckoned  from  a  given  instant.  Determine  the  position  and 
velocity  at  any  time.  Let  distance  be  expressed  in  centimeters  and 
time  in  seconds,  and  assume  that  the  acceleration  increases  by  4  units 
in  1  sec.  Assume  further  that  the  velocity  is  zero  when  the  ac- 
celeration is  zero,  and  that  x  is  reckoned  from  the  position  of  the 
particle  when  the  acceleration  is  zero.  That  is,  the  four  quantities 
/,  v,  x,  t  are  all  zero  together. 

Ans.  x  e=  f/3,     v  =  2/2,    /  =  \t. 

13.  With  data  as  in  Ex.  12,  where  was  the  particle  and  what  was 
its  velocity  3  sec.  before  the  instant  at  which  the  acceleration  was 
zero? 

14.  If  jit  were  reckoned  from  the  position  of  the  particle  3  sec. 
before  the  acceleration  was  zero,  how  would  the  results  of  Ex.  1 2  be 
changed  ?  (Make  no  change  in  the  origin  of  time. )  Does  this 
change  in  the  origin  of  abscissas  imply  a  different  case  of  motion  ? 


MOTION    IN    A   STRAIGHT    LINE.  I77 

§  3.  Motion  and  Force. 
• 

212.  Laws  of  Motion. —  In  the  foregoing  analysis  of  the  motion 

of  a  particle,  nothing  has  been  said  of  the  laws  in  accordance  with 
which  the  motions  of  bodies  actually  occur. 

By  studying  the  motions  of  bodies  in  nature,  it  is  found  that  the 
motion  of  any  given  body  is  influenced  by  its  relation  to  other 
bodies.  This  is  often  expressed  by  saying  that  one  body  is  ' '  acted 
upon ' '  by  other  bodies.  Such  an  ' '  action ' '  of  one  body  upon 
another,  measured  in  a  particular  way,  is  called  a.  force.     (Art.  32.) 

Newton's  three  laws  of  motion  give  a  concise  statement  of  the 
way  in  which  the  motion  of  a  body  is  influenced  by  forces, — i.  e.,  by 
the  action  of  other  bodies.  A  full  understanding  of  these  laws  re- 
quires a  more  general  analysis  of  motion  than  has  been  given  in  this 
Chapter.  The  present  discussion  of  the  laws  of  motion  must  be 
limited  to  the  case  of  motion  in  a  straight  line.  A  formal  statement 
of  Newton' s  laws,  with  a  fuller  explanation  of  their  meaning,  will  be 
given  in  Chapter  XIV. 

213.  Motion  of  a  Body  Which  Is  Not  Influenced  by  Other 
Bodies. —  Newton's  first  law  asserts  that  a  body  not  acted  upon  by 
force  (that  is,  wholly  uninfluenced  by  other  bodies)  will  remain  at 
rest,  or  else  will  move  in  a  straight  line  with  uniform  velocity. 

The  meaning  of  this  law  has  already  been  briefly  explained 
(Art.  31). 

214.  Effect  of  Constant  Force  on  Body  Initially  at  Rest. —  If 
a  force  of  constant  magnitude  and  direction  acts,  for  a  certain  interval 
of  time,  upon  a  body  initially  at  rest,  the  body  will  have  at  the  end 
of  the  interval  a  velocity  whose  direction  is  that  of  the  force,  and 
whose  magnitude  is  proportional  directly  to  the  force  and  to  the 
duration  of  the  interval,  and  inversely  to  the  mass  of  the  body. 

Let  P  denote  the  magnitude  of  the  force,  /  the  duration  of  the 
interval,  and  m  the  mass  of  the  body  acted  upon  ;  then  at  the  end  of 
the  interval  the  velocity  of  the  body  is  proportional  directly  to  P  and 
to  /,  and  inversely  to  m  ;  that  is,  it  is  proportional  to  Pt/m. 

Thus,  if  a  force  P\  acting  for  a  time  t*  upon  a  particle  of  mass 
m  initially  at  rest,  gives  it  a  velocity  v'\  and  if  a  force  P"  acting  for 
a  time  t"  upon  a  particle  of  mass  m   initially  at  rest  gives  it  a  velocity 

*'"'  then  v':v"  =  P't'lm  :P"t"lm". 


178  theoretical  mechanics. 

Examples. 

1.  A  given  force  acting  upon  a  mass  of  1  lb.  for  1  sec.  gives  it  a 
velocity  of  10  ft.-per-sec.  What  velocity  would  an  equal  force  im- 
part to  a  mass  of  5  lbs.  in  4  min.? 

2.  A  force  of  magnitude  P  acting  for  4  sec.  upon  a  body  of  mass 
20  lbs.  gives  it  a  velocity  of  10  ft.-per-sec.  What  velocity  will  be 
imparted  to  a  body  of  50  lbs.  mass  in  9  sec.  by  a  force  of  magnitude 

3.  A  force  of  8  lbs.  (z.  e.,  equal  to  the  weight  of  8  lbs.  mass  ;  see 
Art.  47),  acting  upon  a  body  of  mass  m  lbs.  for  12  sec,  gives  it  a 
velocity  of  45  ft.  -per-sec.  What  velocity  will  be  imparted  to  a  body 
of  mass  7m  lbs.  by  a  force  of  22  lbs.  acting  for  9  sec.  ? 

4.  Two  bodies  whose  masses  are  2  lbs.  and  5  lbs. ,  starting  from 
rest  at  the  same  instant,  are  observed  to  have  equal  velocities  at  every 
subsequent  instant.  Compare  the  magnitudes  of  the  forces  acting 
upon  them. 

5.  A  body  near  the  surface  of  the  earth  will,  if  unsupported,  fall 
toward  the  earth.  It  is  observed  that  two  bodies  of  different  masses, 
starting  from  rest,  acquire  equal  velocities  in  equal  times.  What 
inference  can  be  drawn  as  to  the  forces  acting  upon  them  ? 

6.  A  body  falling  from  rest  under  the  attraction  of  the  earth  is 
observed  to  have  a  velocity  of  32. 2  ft.-per-sec.  at  the  end  of  the  first 
second.  Assuming  the  attraction  of  the  earth  to  be  a  constant  force, 
what  velocity  will  the  body  have  at  the  end  of  20  sec.  ? 

7.  If  g  ft.  -per-sec.  is  the  velocity  acquired  by  a  falling  body  in  1 
sec,  what  is  its  velocity  at  the  end  of  t  sec,  assuming  that  the  only 
force  acting  upon  the  body  is  the  constant  attraction  of  the  earth  ? 

215.  Effect  of  Constant  Force  on  Body  Not  Initially  at  Rest. — 
If  a  body  moving  in  a  straight  line  is  acted  upon  by  a  force  of  con- 
stant magnitude  whose  direction  coincides  with  that  in  which  the 
body  is  moving,  its  velocity  will  receive,  during  any  interval  of  time, 
an  increment  proportional  directly  to  the  force  and  to  the  duration 
of  the  interval,  and  inversely  to  the  mass  of  the  body. 

Let  m  be  the  mass  of  the  body,  P  the  magnitude  of  the  force,  and 
let  the  velocity  have  values  vx  and  vt  at  any  two  instants  tx ,  ty     Then 

v.2  —  vx  is  proportional  to  P  (72  —  t^/m. 

This  case  obviously  includes  that  given  in  the  preceding  Article  ; 
for  if  the  initial  velocity  is  zero,  the  final  velocity  is  equal  to  the 
velocity-increment. 

If  a  body  is  acted  upon  by  a  constant  force  whose  direction  is 
opposite  to  that  in  which  the  body  is  moving,  the  velocity  is  de- 


MOTION    IN    A    STRAIGHT    LINE.  1 79 

creased,  during  any  interval  of  time,  by  an  amount  proportional 
directly  to  the  force  and  to  the  interval,  and  inversely  to  the  mass. 
If  the  interval  be  long  enough,  the  velocity  will  be  reduced  to  zero, 
and  will  then  be  reversed  in  direction.  This  case  and  the  preceding 
are  both  included  in  the  general  statement  that 

A  constant  force  acting  upon  any  body  gives  it  a  velocity -incre- 
ment zvhose  direction  is  that  of  the  force,  and  whose  magnitude  is 
proportional  directly  to  the  force  and  to  the  time  during  which  it  acts, 
and  inversely  to  the  mass  of  the  body. 

If  the  two  directions  along  the  line  of  motion  are  distinguished  as 
plus  and  minus,  the  sign  of  the  velocity-increment  agrees  with  that  of 
the  force. 

It  should  be  observed  that  this  principle  has  no  reference  to  the 
distance  described  by  the  body  during  the  interval  in  question.  It 
gives  us  a  rule  for  estimating  the  change  in  the  velocity  ;  the  result 
is  independent  of  the  actual  velocity  at  the  beginning  of  the  interval 
or  at  any  other  instant.  The  distance  passed  over  depends  upon  the 
value  of  the  velocity  at  every  instant  throughout  the  interval,  not 
merely  upon  the  amount  by  which  the  velocity  changes. 

The  above  proposition  is  a  statement  of  Newton's  second  law  of 
motion,  as  applied  to  the  motion  of  a  body  under  the  action  of  a 
single  force  whose  line  of  action  coincides  with  the  line  of  motion. 
Newton's  third  law  is  the  law  of  action  and  reaction  (Art.  35).  For 
a  formal  statement  of  the  three  laws,  see  Art.  259. 

216.  Equation  of  Motion  for  Particle  Acted  Upon  by  Constant 
Force. —  Let  a  particle  of  mass  m}  acted  upon  for  a  time  At  by  a 
force  P,  receive  a  velocity-increment  Av ;  and  let  a  particle  of  mass 
m\  acted  upon  by  a  force  P'y  receive  in  a  time  At'  a  velocity-incre- 
ment Av'.     Then 

Av  :  Av'  =  PAt/m  :  P'At'/m  ; 

or  mAvjPAt  =  m'Av'/P'At'. 

That  is,  the  quantity  mAv/PAt  has  the  same  value  whatever  the 
particular  case  of  motion  considered  ;  the  force  P  and  the  mass  m 
having  any  values  whatever,  and  Av  being  the  increment  of  velocity 
received  in  a  time  At.     There  may  therefore  be  written  the  equation 

m  Av\PAt  =±  k} 
or  Av  =k(PAtjm),      .         .         .         .     (1) 


l8o  THEORETICAL    MECHANICS. 

k  being  a  constant.  This  may  be  called  the  general  equation  of 
motion  for  a  particle  acted  upon  by  a  constant  force  directed  along 
the  line  of  motion. 

\ip  is  the  acceleration,  the  equation  may  be  written 

p  =  k(Plm)t  ....     (2) 

since  p  =  Av/At  when  the  velocity  varies  at  a  uniform  rate. 

The  numerical  value  of  k  depends  upon  the  units  employed  in 
expressing  the  values  of  Av,  P,  m  and  At.  These  units  having  been 
chosen,  k  can  be  determined  by  a  single  experiment.  The  nature  of 
the  experiment  must  be  as  follows  : 

A  force  of  known  magnitude  (say  P'  units)  is  applied  to  a  body 
of  known  mass  {m'  units)  for  a  known  time  (  At'  units),  and  the  ve- 
locity produced  is  measured  (call  it  Av'  ft.-per-sec).  Then,  since 
these  values  must  satisfy  equation  (i), 

k  =  m'Av'/P'At'. 

The  units  of  force,  mass,  length  and  time  may  thus  be  chosen  arbi- 
trarily and  a  value  of  k  determined  which  will  make  equation  (i) 
true  for  all  cases  in  which  the  same  units  are  employed. 

The  unit  velocity  is  here  assumed  to  be  derived  from  the  units  of 
length  and  of  time  as  in  Art.  191. 

Examples. 

1.  Take  the  unit  mass  as  a  pound,  the  unit  force  as  the  weight  of 
a  pound-mass  {i.  e.y  a  pound-force),  the  unit  time  as  the  second  and 
the  unit  length  as  the  foot.     Determine  the  value  of  k. 

Let  a  body  of  m  pounds-mass  fall  freely  from  rest  under  the  ac- 
tion of  gravity.  Experiment  shows  that  the  velocity  increases  at  a 
uniform  rate.  In  one  second  the  increment  of  velocity  is  about  32.2 
ft.-per-sec.  The  force  producing  this  effect  is  the  weight  of  m 
pounds-mass,  or  in  terms  of  the  pound-force  its  value  is  m.  Hence 
in  the  above  value  of  k  we  may  substitute  m'  =  m,  P'  =  m,  At'  = 
1,  Av'  =  32.2; 

k  =  {m  X  2t2-2)/(.m  X  1)  =  32.2. 
The  equation  of  motion  for  this  system  of  units  is  therefore 
p  =  Av/At  =  2>2.2P/m. 

The  value  32.2  ft. -per-sec. -per-sec.  is  only  an  approximate  value 
of  the  acceleration  of  a  body  falling  freely  under  gravity.  The  true 
value  varies  somewhat  with  the  position  on  the  earth's  surface.  If 
g  denotes  the  value  at  any  given  locality,  k  =  g,  and  the  equation  of 
motion  is  p  =  g<J>\m).         .         .         .         .     (3) 


MOTION    IN    A    STRAIGHT    LINE  l8l 

Since  the  pound-force  varies  with  the  locality  in  exactly  the  same 
ratio  as  the  value  of  gy  the  numerical  value  of  any  given  force  (rep- 
resented by  P  in  the  equation)  varies  inversely  as^-,  so  that  equa- 
tion (3)  is  always  true  if  g  is  given  its  true  value  for  the  particular 
locality  at  which  the  pound-force  is  determined. 

2.  If  the  unit  mass  is  equal  to  200  lbs.,  the  unit  force  equal  to 
the  weight  of  10  lbs.,  the  unit  time  the  second,  and  the  unit  length 
the  foot,  determine  the  value  of  k.  Ans.  k  =  g/20. 

3.  If  the  pound  is  the  unit  mass,  the  foot  the  unit  length,  and 
the  second  the  unit  time,  what  must  be  the  unit  force  in  order  that 
k  may  equal  1  ? 

Ans.  A  force  equal  to  i/g  times  the  weight  of  1  lb. 

4.  If  the  pound-force  is  the  unit  force,  the  foot  the  unit  length, 
and  the  second  the  unit  time,  what  must  be  the  unit  mass  in  order 
that  k  may  equal  1  ?  Ans.  A  mass  equal  to  g  lbs. 

5.  Show  that,  whatever  units  are  employed,  the  constant  k  is  nu- 
merically equal  to  the  acceleration  due  to  the  unit  force  acting  upon 
the  unit  mass.  Answer  examples  1,  2,  3  and  4  by  the  direct  appli- 
cation of  this  general  result. 

6.  The  equation  of  motion  may  be  written  P  =  k'mp,  in  which 
k'  =  \\k.  Show  that  the  constant  k'  is  numerically  equal  to  the 
force  which  will  give  the  unit  mass  the  unit  acceleration. 

217.  Kinetic  Systems  of  Units. —  It  has  been  seen  that,  in  the 
general  equation  of  motion  given  above  (Art.  216),  the  four  quanti- 
ties, force,  mass,  velocity  and  time  (or  force,  mass,  length  and  time) 
may  be  expressed  in  any  arbitrary  units,  provided  the  value  of  k  is 
properly  determined.  It  is  also  apparent  that  k  may  be  given  any 
desired  value  by  properly  choosing  the  units  of  force,  mass,  length 
and  time. 

In  order  to  simplify  the  equation  of  motion,  let  k  =  1,  and  con- 
sider what  restriction  is  thus  imposed  upon  the  choice  of  units. 

The  equation  of  motion  becomes 

p  =  AvjAt  =  Pirn.  .         .         .     (1) 

In  order  to  satisfy  this  equation  it  is  necessary  to  express  force, 
mass,  length  and  time  in  such  units  that  a  unit  force  acting  for  a 
unit  time  upon  a  unit  mass  will  give  it  a  unit  velocity. 

Any  system  of  units  satisfying  this  requirement  may  be  called  a 
kinetic  system. 

It  is  obvious  that,  even  with  this  restriction,  any  three  of  the  four 
units  named  may  be  chosen  arbitrarily.  But  when  three  are  chosen 
as  fundamental,  the  fourth  becomes  a  derived  unit. 


1 82  THEORETICAL    MECHANICS. 

For  the  purposes  of  pure  science  the  common  practice  is  to  take 
as  fundamental  the  units  of  mass,  length  and  time ;  the  unit  force 
being  derived  from  these  in  accordance  with  the  requirement  above 
stated.     Two  such  systems  of  units  may  be  mentioned. 

The  centimeter -gram- second  system. — In  this  system  the  cent- 
imeter is  the  unit  length,  the  gram  the  unit  mass  and  the  second  the 
unit  time.  It  is  briefly  called  the  C.  G.  S.  system.  The  unit  force 
is  called  a  dyne. 

A  dyne  is  a  force  which,  acting  for  one  second  upon  a  mass  of 
one  gram,  gives  it  a  velocity  of  one  centimeter-per-second. 

The  C.  G.  S.  system  is  almost  universally  employed  in  purely 
scientific  work. 

The  foot-pound-second  system. — In  this  system  (called  briefly  the 
F.  P.  S.  system)  the  foot  is  the  unit  length,  the  pound  the  unit  mass 
and  the  second  the  unit  time.     The  unit  force  is  called  a  poundal. 

A  poundal  is  a  force  which,  acting  for  one  second  upon  a  mass  of 
one  pound,  gives  it  a  velocity  of  one  foot-per-second. 

Examples. 

i.  A  body  whose  mass  is  40  lbs.  is  acted  upon  by  a  constant 
force  of  12  poundals.  What  is  the  velocity  after  4  sec.  (a)  if  in- 
itially at  rest,  (b)  if  the  initial  velocity  is  20  ft.-per-sec.  in  the  direction 
of  the  force,  (V)  if  the  initial  velocity  is  20  ft.-per-sec.  in  the  direction 
opposite  to  that  of  the  force. 

Ans.  {a)  1.2  ft.-per-sec.  (b)  21.2  ft.-per-sec.  (c)  18.8  ft.-per- 
sec. 

2.  A  body  of  20  lbs.  mass  is  acted  upon  by  a  constant  force  which 
in  1 2  sec.  gives  it  a  velocity  of  60  ft.  -per-sec.  What  is  the  mag- 
nitude of  the  force  in  poundals? 

3.  A  body  of  6  lbs.  mass,  starting  from  rest  and  falling  freely  under 
the  earth' s  attraction,  is  observed  to  have,  after  2  sec. ,  a  velocity  of 
64. 4  ft.  -per-sec.  If  the  earth' s  attraction  upon  the  body  is  a  constant 
force,  what  is  its  magnitude  in  poundals  ? 

4.  A  body  of  m  lbs.  mass,  falling  vertically  under  the  action  of 
its  own  weight,  receives  during  each  second  a  velocity  of  g  ft.  -per-sec. 
What  is  its  weight  in  poundals  ?  (That  is,  what  attractive  force  is 
exerted  upon  it  by  the  earth?)  Ans.  mg  poundals. 

5.  What  is  the  ratio  between  a  force  of  one  pound  and  a  force  of 
one  poundal? 

6.  A  body  of  20  gr.  mass  is  acted  upon  by  a  constant  force  of  5 
dynes.      Determine  its  velocity  after   16  sec.  {a)  if  initially  at  rest, 


MOTION    IN    A   STRAIGHT    LINE.  1 83 

(b)  if  its  initial  velocity  is  16  c.m.-per-sec.  in  the  direction  of  the 
force,  (c)  if  its  initial  velocity  is  16  c.m.-per-sec.  in  the  direction 
opposite  to  that  of  the  force.  Ans.   (c)  12  c.m.-per-sec. 

7.  A  body  whose  mass  is  3  kilogr.  is  acted  upon  by  a  constant 
force  which,  in  5  sec,  changes  its  velocity  from  10  met. -per-sec.  in 
one  direction  to  12  met. -per-sec.  in  the  opposite  direction.  What  is 
the  value  of  the  force  ? 

Ans.    1,320,000  dynes  in  the  direction  of  the  final  velocity. 

8.  A  body  of  2,000  gr.  mass,  starting  from  rest  and  falling  freely 
under  the  earth's  attraction,  has  at  the  end  of  2  sec.  a  velocity  of 
19.6  met. -per-sec.  If  the  earth's  attraction  upon  the  body  is  a  con- 
stant force,  what  is  its  magnitude  in  dynes  ? 

Ans.  1.96  X  1  o6  dynes. 

218.  Engineers'  Kinetic  System. —  For  the  purposes  of  the  en- 
gineer the  most  convenient  unit  of  force  is  the  weight  of  a  definite 
mass  at  the  earth's  surface.  Such  a  unit,  though  not  exactly  the 
same  for  all  localities,  is  sufficiently  definite  for  most  practical  pur- 
poses. 

Let  the  pound-force  (already  defined  as  the  weight  of  a  pound- 
mass)  be  taken  as  the  unit  force,  let  the  foot  and  second  be  taken  as 
units  of  length  and  time,  and  let  the  unit  mass  be  so  determined  as 
to  satisfy  the  equation  P  =  m(Av/Af)  =  mp. 

The  unit  mass  must  now  be  defined  as  a  mass  whose  velocity 
will  increase  by  one  foot-per-second  during  each  second  if  acted 
tip  on  by  a  force  of  one  pound . 

The  ratio  of  this  unit  to  the  pound-mass  may  readily  be  deter- 
mined. A  force  of  1  lb.  acting  upon  a  mass  of  1  lb.  for  1  sec.  gives 
it  a  velocity  of^-  ft.  -per-sec.  The  same  force  acting  upon  a  mass  of 
g  lbs.  for  1  sec.  will  therefore  give  it  a  velocity  of  1  ft.  -per-sec.  The 
required  unit  mass  is  therefore  g  times  as  great  as  the  pound-mass. 

If  this  system  of  units  is  employed,  a  mass  whose  value  is  given 
in  pounds  must  be  reduced  to  the  unit  just  defined  by  dividing 

It  will  hereafter  be  assumed,  unless  otherwise  expressly  stated, 
that  a  kinetic  system  of  units  is  employed,  so  that  the  equation  of 
motion  takes  the  form 

P=  mp. 

Examples. 
1.   A  mass  of  half  a  ton  is  acted   upon   by  a  force  of  50  lbs. 


184  THEORETICAL    MECHANICS. 

Write  the  equation  of  motion,  using  the  pound  as  the  unit  force. 
What  is  the  acceleration  ?     [1  ton  =  2,000  lbs.] 

Ans.  p  =  1 .61  ft. -per-sec. -per-sec. 

2.  With  French  units,  let  the  unit  length  be  the  meter,  the  unit 
time  the  second  and  the  unit  force  the  weight  of  a  kilogram.  Deter- 
mine the  value  of  the  unit  mass  in  terms  of  the  kilogram,  so  that  the 
equation  P  =  nip  may  be  satisfied. 

3.  A  mass  of  200  kilograms  is  acted  upon  by  a  force  equal  to 
half  its  weight.  Write  the  equation  of  motion,  taking  units  as  in 
Ex.  2.     Determine  the  acceleration. 

Ans.  p  =  4.9  met. -per-sec. -per-sec. 

219.  Dimensions  of  Units  in  Kinetic  System. —  The  relation 
which  must  subsist  among  the  units  in  order  that  the  constant  k  in 
the  general  equation  of  motion  shall  be  unity  may  be  expressed  by 
a  dimensional  equation.  Let  the  units  be  represented  by  the  same 
symbols  as  heretofore  (Arts.  15,  191,  204). 

The  equation  to  be  satisfied  is  P  =  mp.  The  dimensional  equa- 
tion is  therefore  p  __  j^ 

But  also 

V  =  L/T  ;     A  =  V/T  =  L/T2; 
hence 

F  =  ML/T2. 

If,  as  in  the  C.  G.  S.  and  the  F.  P.  S.  systems  (Art.  217),  the 
fundamental  units  are  M,  L  and  T,  this  equation  gives  the  dimen- 
sions of  the  unit  force.  If,  however,  F,  L  and  T  are  made  funda- 
mental, the  equation  gives  the  dimensions  of  the  unit  mass  ;  in  this 

case  it  may  be  written 

M  =  FT2/L. 

In  accordance  with  scientific  usage,  we  shall  generally  regard  M, 
L  and  T  as  fundamental  units  in  the  general  equation  of  motion  and 
all  equations  derived  from  it ;  force  being  regarded  as  having  the 
dimensions  given  by  the  above  equation. 

220.  Effect  of  Two  or  More  Constant  Forces. —  If  a  body  mov- 
ing in  a  straight  line  be  acted  upon  by  two  or  more  forces  directed 
along  the  line  of  motion,  it  receives,  during  any  interval,  a  velocity- 
increment  equal  to  the  algebraic  sum  of  the  increments  which  would 
be  produced  by  the  forces  acting  separately.  The  same  effect  would 
be  produced  by  a  single  force  equal  to  the  algebraic  sum  of  the  sev- 
eral forces.     That  is,  the  resultant  of  any  number  of  forces  acting 


MOTION    IN   A   STRAIGHT    LINE.  1 85 

upon  the  same  particle  and  directed  along  its  line  of  motion  is  a 
single  force  equal  to  their  algebraic  sum. 

221.  Effect  of  Variable  Force. —  If  a  body  is  acted  upon  by  a 
force  of  variable  magnitude,  the  velocity-increment  produced  in  any 
given  interval  cannot  be  estimated  so  simply  as  in  the  case  of  a  con- 
stant force.  In  this  case  the  equation  of  motion  becomes  a  differential 
equation. 

If  a  body  of  mass  m  receives  a  velocity-increment  Av  in  a  time 
A/,  the  equation 

P  =  m(Av/At) 

gives  the  magnitude  of  a  constant  force  which  will  produce  the  same 
velocity-increment  in  the  same  time.  If  the  interval  At  is  long,  the 
actual  magnitude  of  the  force  may  vary  widely  from  this  value.  But 
if  the  time  At  be  taken  smaller  and  smaller,  approaching  the  limit 
zero,  Av  also  approaching  zero,  the  value  of  P  given  by  the  equation 
approaches  as  a  limit  the  actual  magnitude  of  the  force  at  the  begin- 
ning of  the  interval.  Hence,  if  P  denotes  the  value  of  the  force  at 
any  instant, 

P  =  lim  [m(Av/Af)]  =  m{dvjdf). 

The  determination  of  the  change  of  velocity  produced  by  the  force 
in  any  finite  time  requires  the  integration  of  this  equation.  This  is 
not  possible  unless  the  law  of  variation  of  P  is  known. 

222.  Equation  of  Motion  of  a  Particle  Acted  Upon  by  Any 
Number  of  Constant  or  Variable  Forces. —  It  is  now  evident  that, 
if  a  kinetic  system  of  units  be  employed,  the  equation  of  motion  may 
be  written 

P  =  m{dvjdt),     or     P=  mp, 

if p  is  the  instantaneous  value  of  the  acceleration  and  P  the  instan- 
taneous value  of  the  algebraic  sum  of  all  forces  acting  on  the  particle. 
If  the  position  of  the  particle  be  specified  by  its  distance  x  from  a 
fixed  point  in  the  line  of  motion,  p  =  dvjdt  =  d2xldt'\  and  the 
equation  of  motion  may  be  written 

P  =  m(d2x/dt2). 


CHAPTER    XIII. 

MOTION    IN    A    STRAIGHT    LINE  :    APPLICATIONS. 

§  i.   General  Method. 

223.  Classes  of  Problems. — The  equation 

P=mp  .         .         .         .     (1) 

may  always  be  applied  in  the  solution  of  problems  relating  to  the 
motion  of  a  particle  in  a  straight  line.  The  problems  that  may  con- 
ceivably arise  are  of  various  kinds,  depending  upon  what  is  known 
and  what  unknown  regarding  the  motion.  The  most  important  cases 
are  the  following  : 

(1)  The  motion  being  known,  it  is  required  to  determine  the  re- 
sultant force. 

(2)  The  forces  being  known  (as  functions  of  the  position  or  of 
the  time  or  of  both),  it  is  required  to  determine  the  motion. 

The  second  of  these  problems  will  require  the  solution  of  a  dif- 
ferential equation,  and  may  be  called  the  inverse  problem  ;  the  first 
will  be  called  the  direct  problem,  since  it  involves  no  integration. 

224.  Direct  Problem :  To  Determine  the  Resultant  Force  When 
the  Motion  Is  Known. — If  the  acceleration  is  known  at  every  in- 
stant, the  resultant  force  can  be  determined  by  substitution  in  the 
equation  P  =  mp.  If  the  position  or  the  velocity  is  known  as  a 
function  of  the  time,  the  acceleration  can  be  found  by  differentiation. 

Examples. 

1.  A  body  of  1  lb.  mass,  starting  from  rest,  moves  so  that  its  dis- 
tance from  the  starting  point  at  every  instant  is  given  by  the  formula 
x  =  16. 1 1'\  x  being  in  feet  and  t  in  seconds.  Required  the  mag- 
nitude of  the  resultant  force  acting  on  the  body  at  any  instant. 

Ans.  The  force  is  constant  and  equal  to  32.2  poundals. 

2.  If  the  velocity  of  a  body  is  constant,  what  is  the  magnitude  of 
the  resultant  force  acting  on  it  ? 

3.  If  the  position  of  a  particle  of  mass  m  is  given  by  the  formula 
x  =  a  sin  bt,  determine  the  value  of  the  resultant  force  acting  upon 
it  as  a  function  of  x.  Ans.  P  =  — mblx. 


MOTION    IN    A    STRAIGHT    LINE:    APPLICATIONS.  1 87 

4.  A  body  of  m  lbs.  mass,  acted  upon  by  no  force  except  the 
earth's  attraction,  is  observed  to  receive  each  second  a  velocity- 
increment  of  g  ft.  -per-sec.  What  is  the  magnitude  of  the  force  act- 
ing upon  it  ?  Ans.  mg  poundals. 

5.  A  body  whose  mass  is  18  lbs.  moves  so  that  its  position  at 
any  instant  is  given  by  the  equation  x  =  $t2  -j-  6/  -f  8,  /  being  in 
seconds  and  x  in  feet.  Required  the  magnitude  of  the  resultant 
force  acting  upon  it  at  any  instant. 

Ans.    180  poundals  or  5.59  pounds-force. 

6.  What  force  will  give  an  acceleration  of  1,000  cm. -per-sec  - 
per.  -sec.  to  a  mass  of  600  gr.  ? 

7.  What  constant  force  acting  upon  a  particle  of  m  grams  mass 
will  increase  its  velocity  by^  cm. -per-sec.  in  1  sec.  ? 

8.  What  is  the  weight  in  dynes  of  a  mass  of  m  grams  ? 

Ans.  mg,  if  g  is  the  acceleration  due  to  gravity,  expressed  in 
cm. -per-sec. -per-sec.     Its  value  is  known  to  be  about  981. 

9.  The  velocity  of  a  particle  is  proportional  to  its  distance  from 
a  fixed  point,  and  is  24  ft. -per-sec  when  the  distance  from  the  fixed 
point  is  2  ft.  If  the  mass  of  the  particle  is  4  lbs.,  what  is  the  value 
of  the  resultant  force  acting  upon  it  when  8  ft.  from  the  fixed  point? 
Also  when  3  ft.  from  the  fixed  point  ? 

Ans.   4,608  poundals.      1 ,728  poundals. 

225.  Inverse  Problem:  To  Determine  the  Motion  When  the 
Forces  Are  Known. —  If  every  force  is  known  as  a  function  of  one  or 
more  of  the  three  quantities  x,  t,  v,  the  general  equation  P  =  mp 
becomes  a  differential  equation,  the  complete  solution  of  which  de- 
termines x  and  v  as  functions  of  t. 

In  general  two  integrations  will  be  required,  thus  introducing  two 
constants.  To  determine  the  constants,  some  information  must  be 
given  concerning  the  motion  at  one  or  more  definite  instants,  or  in 
some  one  or  more  definite  positions.  The  information  usually  avail- 
able is  the  following :  The  velocity  and  position  at  a  certain  instant 
are  completely  known. 

Note  on  constants  of  integration. — The  method  of  determining 
constants  of  integration  has  already  been  illustrated  in  several  par- 
ticular cases.  A  clear  understanding  of  the  general  method  is  of 
importance  to  the  student.  To  determine  such  a  constant  it  is  always 
necessary  to  have  some  information  in  addition  to  that  which  enables 
us  to  write  the  differential  equation. 

Let  there  be  given  any  equation  containing  two  variables,  x  and 
yy  and  a  constant,  C.     Then  the  value  of  C  can  be  determined  if  one 


1 88  THEORETICAL    MECHANICS. 

pair  of  simultaneous  values  of;r  and  jy  be  known  ;  that  is,  if  it  be  known 
that  ' '  when  x  =  some  known  value,  y  =  some  known  value. ' ' 
Thus,  let  the  given  equation  be 

ffcy,  C)  =  o. 

If  it  is  known  that  when  x  =  a,  y  =  b,  a  and  b  being  known  con- 
stants, there  may  be  written 

/{a,  b,  C)  =  o, 

from  which  C  can  be  determined. 

In  a  problem  relating  to  the  motion  of  a  particle,  the  variables 
being  usually  some  two  of  the  quantities  x,  t  and  v,  the  information 
necessary  for  determining  the  constant  is  equivalent  to  some  knowl- 
edge as  to  the  position  at  some  instant,  or  as  to  the  condition  of 
motion  of  the  particle  either  at  some  instant  or  in  some  position. 

It  is  to  be  noticed  that  the  same  method  will  serve  for  determining 
any  constant  in  an  equation,  whether  introduced  by  integration  or 
otherwise. 

§  2.   Motion  Under  Constant  Force. 

226.  Solution  of  General  Problem.—  Let  a  particle  of  mass  m 

be  acted  upon  by  forces  whose  resultant 

n jc      •  has  the  constant  value  P  directed  parallel 


O  A.         to  the  line  of  motion.     Let  O  (Fig.  103) 

Fig.  103.  be  a  fixed  point  in  the  line  of  motion,  and 

let  x  be  the  distance  of  the  particle  from 
O  at  the  time  t.     The  equation  of  motion  is  then 

mOPxfdt*)  =  P  =  constant.     .         .         .     (1) 

Two  integrations  give 

midxldf)  =Pt+ Cx\         .         .         .     (2) 

m*:==iFP'+)CS+.CM.      .        .        .     (3) 

To  detemine  CA  and  C.2,  let  it  be  known  that  at  a  certain  instant  t0 
the  velocity  is  vQ  and  the  abscissa  of  the  particle  x0.     Then  from  (2), 

and  from  (3),  Cx  =  mva~Pk; 

C2  =  mx„  —  iP/0'  —  (ptv,  —  P/0)l0  =  mx„  —  mvja  +  h_P/„\ 


MOTION    IN   A   STRAIGHT    LINE:    APPLICATIONS.  189 

Since  the  origin  of  time  may  be  chosen  arbitrarily,  let  t  be  reckoned 
from  the  instant  at  which  v  =  vQ  and  x  =  x0;  then  t0  =  o,  and 

Cx  —  mv0 ;     C2  =  mx0. 

Equations  (2)  and  (3)  now  become 

midxjdf)  =  Pt  -f  mv0 ;     or     m{v  —  v0)  =  Pt\      .     (4) 

*&r  =  \Pf  4  w%/  +  **#o-      •        •        •    (5) 

These  equations  give  the  velocity  and  position  at  any  instant. 

227.  Motion  of  a  Body  Falling  Vertically  Near  the  Earth.— 
A  body  near  the  surface  of  the  earth  is  attracted  toward  the  earth's 
center  with  a  force  which  varies  as  the  body  moves  upward  or  down- 
ward. If  the  range  of  motion  is  small  compared  with  the  earth's 
radius,  the  force  varies  so  little  that  for  most  purposes  it  may  be 
regarded  as  constant.  Hence  if  a  body  starts  from  rest,  or  is  pro- 
jected vertically  upward  or  downward,  and  is  then  left  to  the  influence 
of  the  earth's  attraction,  it  presents  a  case  of  rectilinear  motion  under 
a  constant  force. 

Experience  shows  that  bodies  of  unequal  mass,  acted  upon  by 
gravity  alone,  are  equally  accelerated.  Their  weights  are  therefore 
proportional  to  their  masses.  Let  Wx  and  W2  denote  the  weights 
of  two  bodies  whose  masses  are  mx  and  m2 ;  then 

WJIV2  =  mjm2. 

This  equation  is  true  for  any  units  of  force  and  mass.  If  these  units 
satisfy  the  condition  prescribed  in  Art.  217  (so  that  the  unit  force 
gives  the  unit  mass  the  unit  acceleration),  the  acceleration  of  the 
mass  mx  acted  upon  by  the  force  Wx  is  WJmx ,  and  the  acceleration 
of  the  mass  m2  acted  upon  by  the  force  W2  is  WJm2.  If  the  known 
value  *  of  the  acceleration  due  to  gravity  is  denoted  by  g, 

*The  value  of  g  varies  with  the  position  on  the  earth.  This  variation  is 
approximately  represented  by  the  following  formula :  Let  <p  denote  the  lat- 
itude and  h  (centimeters)  the  elevation  above  sea-level.  Then  in  C.  G.  S. 
units  (c.m.-per-sec.-per-sec), 

g  =  980.6056  —  2.5028  cos  2<p  —  0.000003/!. 

In  the  numerical  exercises  in  this  book  in  which  British  units  are  em- 
ployed the  value  32.2  ft.-per-sec.-per-sec.  may  be  used  ;  in  French  units  the 
value  981  c.m.-per-sec.-per-sec.  is  sufficiently  correct. 

The  effect  of  the  earth's  rotation  on  the  value  of  g  is  considered  in 
Art.  311. 


I90  THEORETICAL    MECHANICS. 

WJm^  =  WJm.z  =  g. 

Or,  if   W  is  the  weight  of  any  body  of  mass  tn, 

W\m  =  g. 

This  equation  may  also  be  written 

W  =  mg,     or     m  =  W[g. 

Thus,  in  any  equation  in  which  kinetic  units  are  employed,  the 
weight  of  a  body  of  mass  m  may  be  put  equal  to  mg. 
The  equation  of  motion 

P  =  mp, 

applied  to  the  case  of  a  particle  of  mass  m  acted  upon  by  no  force 
except  its  weight,  becomes 

mg  =  mp,     or    p  =  g. 
This  may  be  written 

d^x/dt2  =  g  =  constant,         .         .         .     (1) 

if  x  denotes  the  distance  of  the  particle  from  a  fixed  point  in  the  line 
of  motion.      The  positive  direction  for  x  is  downward. 

Let  equation  (1)  be  integrated,  and  let  the  conditions  for  deter- 
mining the  constants  of  integration  be  that  when  /  =  o,  x  =  x0  and 
V  =  v0.     The  first  integration  gives 

dxjdt  =  gt  -f  7'o ;  .  .  .      (2) 

and  the  second  gives 

*=  kg**  +  Vot  +  *o.  •  •  •       (3) 

These  results  might  have  been  deduced  immediately  from  equa- 
tions (4)  and  (5),  Art.  226,  by  substituting  g  for  P/m. 

Consider  the  three  cases  in  which  the  initial  velocity  is  zero,  pos- 
itive, and  negative,  respectively. 

(1)  Let  the  body  fall  from  rest,  and  take  the  starting  point  as  the 
origin  for  reckoning  x,  t  being  reckoned  from  the  instant  when  the 
body  is  at  the  origin.  Then  v0  —  o,  and  equations  (2)  and  (3)  be- 
come 

dx/dt  =  v  =  gt;  .         .  (4) 

x=Wl (5) 

(2)  If  the  body  has  initially  a  velocity  downward,  the  general  for- 
mulas (2)  and  (3)  apply,  v0  having  a  positive  value.  Let  vQ  =  V, 
and  let  x  be  measured  from  the  point  of  projection,  so  that  x0  =  o. 


MOTION    IN    A   STRAIGHT    LINE:    APPLICATIONS.  191 

Then                               dx/dt  =  v  =  gt  +  V;          .         .         .     (6) 
x  -  Jjtf  -f  *? (7) 

(3)  If  the  body  is  given  an  initial  velocity  V  upward,  and  if  x 
is  reckoned  from  the  initial  position  and  is  positive  downward, 
v0  =  —  V,  x0  =  0,  and  equations  (2)  and  (3)  become 

dx/dt  =  v=gt  —  V\  .         .         .     (8) 

x^\gt2—Vt.      ....     (9) 

Examples. 

1.  Prove  that  the  velocity  v  of  a  body  which  has  fallen  verti- 
cally a  distance  x  from  rest  is  given  by  the  formula  v2  =  2gx. 

2.  Taking  the  value  of  g  in  centimeter-second  units  as  981,  com- 
pute the  velocity  of  a  body  after  falling  10  met.  from  rest. 

3.  A  body  is  projected  upward  with  a  velocity  of  100  ft.-per-sec. 
When  will  it  come  to  rest,  how  high  will  it  rise,  and  when  will  it 
return  to  the  starting  point  ? 

4.  A  body  is  projected  upward  with  a  velocity  of  80  ft.  -per -sec. 
After  what  time  will  it  be  20  ft.  above  the  initial  position  ?  Explain 
the  double  answer.     Take^"  =  32.2  ft. -sec.  units. 

Ans.  After  0.26  sec.  or  4.69  sec. 

5.  A  body  is  projected  upward  with  velocity  V.  Show  that  it 
will  rise  to  a  height  V2J2g;  that  it  will  come  to  rest  after  a  time  V/g; 
and  that  it  will  return  to  the  point  of  projection  after  a  time  2  V/g. 

6.  A  body  is  dropped  into  a  well  84  ft.  deep.  How  long  before 
the  sound  of  striking  the  bottom  will  be  heard,  if  the  velocity  of 
sound  is  1, 100  ft.-per-sec?  Ans.  2.36  sec. 

7.  A  body  is  dropped  into  a  well,  and  the  sound  of  striking  the 
bottom  is  heard  after  4  sec.     How  deep  is  the  well?    Ans.  231  ft. 

8.  A  body  is  projected  upward  with  velocity  V.  Show  that  after 
rising  a  distance  h  its  velocity  is  given  by  the  formula  v2  =  V2  —  2gh. 

9.  If  a  body  is  moving  vertically  under  the  action  of  gravity, 
prove  that  its  average  velocity  during  any  interval  of  time  is  equal  to 
its  velocity  at  the  middle  instant  of  the  interval.  The  same  is  true 
in  any  case  of  constant  acceleration. 

10.  At  a  certain  instant  a  body  (acted  upon  by  gravity  alone)  is 
moving*  upward  at  the  rate  of  10  ft.-per-sec.  What  is  its  average 
velocity  for  the  next  half-second  ?  Determine  its  final  position  by 
means  of  the  average  velocity. 

1 1 .  What  is  the  average  velocity  of  a  falling  body  during  the  nth. 
second  after  starting  from  rest?  Ans.   {n  —  ^^ft.-per-sec. 


I92  THEORETICAL   MECHANICS. 

1 2.  What  distance  (in  centimeters)  is  described  by  a  falling  body 
during  the  5th  second  after  starting  from  rest  ? 

1 3.  An  elevator,  starting  from  rest,  has  a  downward  acceleration 
g/2  for  1  sec. ,  then  moves  uniformly  for  2  sec. ,  then  has  an  upward 
acceleration  gfo  until  it  comes  to  rest,  {a)  How  far  does  it  descend  ? 
(b)  A  person  whose  weight  is  140  lbs.  experiences  what  pressure 
from  the  elevator  during  each  of  the  three  periods  of  its  motion  ? 

Ans.  (a)  13^/8  ft.  (b)  70  pounds-weight ;  140  pounds-weight ; 
186^  pounds-weight. 

14.  A  body  of  m  lbs.  mass  rests  upon  a  horizontal  platform.  If 
the  platform  begins  to  fall  with  acceleration  g,  what  pressure  does  it 
exert  upon  the  body  ?  What  is  the  pressure  if  the  platform  begins 
to  rise  with  acceleration  g? 

15.  In  Ex.  14,  determine  the  pressures  in  the  two  cases  in  which 
the  acceleration  is  2g  upward  and  2g  downward. 

1 6.  Equal  masses  of  ni  lbs.  each  rest  upon  two  platforms,  one  of 
which  has  at  a  certain  instant  a  velocity  of  20  ft.  -per-sec.  upward  and 
the  other  a  velocity  of  20  ft. -per-sec.  downward.  Both  platforms 
have  an  upward  acceleration  g\\.  Compare  the  pressures  of  the 
platforms  on  the  two  bodies. 

17.  Velocity  is  imparted  to  a  body  of  5  lbs.  mass  by  means  of  an 
attached  string  whose  breaking  strength  is  a  pull  of  2  lbs.  How 
great  a  velocity  can  the  body  receive  in  2  sec.  ? 

Ans.   4^/5  ft. -per-sec. 

18.  A  string  which  can  just  sustain  a  mass  of  4  lbs.  against  grav- 
ity is  attached  to  a  body  whose  mass  is  1  lb.  which  rests  upon  a 
smooth  horizontal  plane.  Is  it  possible  to  break  the  string  by  a  hor- 
izontal jerk  ?  How  great  an  acceleration  can  be  given  to  the  body 
by  means  of  the  string? 

19.  A  ball  whose  mass  is  5  oz.  is  moving  at  the  rate  of  100  ft.- 
per-sec.  when  it  receives  a  blow  which  exactly  reverses  its  velocity. 
If  the  force  exerted  upon  the  ball  is  constant  and  acts  for  o.  1  sec. , 
what  is  its  magnitude  ? 

An s.  19.4  pounds-force.  Actually,  the  force  would  increase 
from  o  up  to  a  value  much  greater  than  19.4  lbs.  and  then  decrease 
to  o.  The  average  force  is  19.4  if  the  time  occupied  by  the  blow  is 
o.  1  sec. 

20.  In  a  locality  where  the  value  of  g  is  32.2  ft. -per-sec. -per-sec. 
a  body  of  m  lbs.  mass  falls  15.9  ft.  from  rest  in  one  sec.  What  is  the 
average  value  of  the  resistance  of  the  air  ? 

Ans.  0.0124;;/  pounds-force. 


MOTION    IN    A    STRAIGHT    LINE:    APPLICATIONS.  1 93 

§  3.   Force    Varying  With  Distance  From  a  Fixed  Point. 

228.  General  Problem:   Force  Any  Function  of  Distance. — 

In  dealing  with  the  the  forces  of  nature,  an  important  case  to  be  con- 
sidered is  that  in  which  the  force  acting  upon  a  particle  is  directed 
toward  or  from  a  fixed  point  (or  one  which  may  be  regarded  as 
fixed),  its  magnitude  being  some  function  of  the  distance  of  the  par- 
ticle from  the  point.  This  case  will  now  be  considered,  the  motion 
being  restricted  to  a  straight  line  containing  the  fixed  point. 

The  point  toward  (or  from)  which  the  force  is  directed  may  be 
called  the  center  of  force.  If  the  force  is  directed  toward  the  center 
it  is  called  attractive ;  if  from  the  center, 

repulsive.    For  convenience  let  the  origin  K &      , 

(O,  Fig.  104)  be  taken  at  the  center  of         q  ^l 

force,  and  let  the  force  be  reckoned  as  if  Fig.  104. 

it  were  repulsive  in  all  cases.      Let  P  be 

the  magnitude  of  the  force  for  any  position  of  the  particle  ;  then  the 

general  equation  (Art.  222)  is  m(d'2x/dt2)  =  P,  or 

m(dx/dt)  =  P. 

Since  in  the  present  case  P  is  supposed  to  be  a  function  of  x,  the 
equation  becomes 

m(dx/dt)  =  f{x) (1) 

In  any  special  case  the  form  of  the  function/^)  is  known,  and  the 
solution  of  the  differential  equation  (1)  gives  the  relation  between  x 
and  /,  and  also  the  relation  between  x  (or  v)  and  /,  thus  determining 
the  motion  completely.  The  values  of  x  and  v  will  involve  constants 
of  integration,  to  determine  which  certain  "  initial  conditions"  must 
be  specified. 

In  the  following  Articles  special  cases  of  equation  (1)  will  be  con- 
sidered.  One  important  general  result  may,  however,  be  here  noticed. 

Multiplying  both  members  of  (1)  by  xdt  =  dx,  the  first  member 
becomes  integrable.     Thus,  we  have  first, 

mx  dx  =  f{x)dx. 
Integrating,  r 

imx*  =J  f{x)dx  +  C  =  F(x)  +  C, 

F(x)  being  obtained  by  immediate  integration  as  soon  as  the  form 
of  fix)  is  known.     The  last  equation  shows  that  the  velocity  is  a 

13 


194  THEORETICAL    MECHANICS. 

function  of  the  position  ;  that  is,  if  the  particle  comes  more  than 
once  into  the  same  position,  the  velocity  has  the  same  value,  except 
that  its  direction  may  be  reversed. 

The  quantity  \mx2  or  \mv2  is  an  important  one,  and  is  dis- 
cussed in  Chapter  XVII.  The  name  kinetic  energy  is  given  to  it, 
for  reasons  to  be  explained  in  that  chapter. 

229.  Attractive  Force  Proportional  Directly  to  the  Distance. — 

An  important  case  is  that  in  which  the  magnitude  of  the  force  is  pro- 
portional directly  to  the  distance  of  the  particle  from  the  fixed  point. 
The  force  may  be  either  attractive  or  repulsive.  Consider  first  the 
case  in  which  it  is  attractive,  —  i.  *?.,  directed  always  toward  the  fixed 
point. 

Let  P'  be  the  magnitude  of  the  attractive  force  when  the  particle 
is  at  the  unit  distance  from  O  (Fig.  104)  ;  then  for  any  distance  x 
the  force  is  P  =  — P'x,  and  the  equation  of  motion  becomes 

nix  =  — P'x, 
or  dvjdt  =  x  =  — P'x/m  =  — kx,  .         .     (1) 

if  k  be  written  for  the  attractive  force  per  unit  mass  at  unit  distance 
from  O.  The  reason  for  the  minus  sign  is  that  the  force,  being  always 
directed  toward  the  point  O,  is  always  opposite  in  sign  to  x. 

In  order  to  completely  determine  the  motion,  certain  "initial 
conditions"  must  be  known,  in  addition  to  the  differential  equation 
(1)  which  expresses  a  relation  that  is  satisfied  in  every  position  of 
the  particle.  It  will  be  well  to  state  explicitly  all  the  data  which 
serve  to  determine  the  solution  of  the  problem.  Let  the  following  as- 
sumptions be  made: 

(a)  The  origin  being  at  O,  the  fixed  point  toward  which  the  force 
is  always  directed,  let  the  positive  direction  be  toward  the  right,  so 
that  when  the  particle  is  on  the  right  of  O,  x  is  positive  and  the  force 
negative  ;  and  when  the  particle  is  on  the  left  of  O,  x  is  negative  and 
the  force  positive. 

(b)  Take  the  ' '  origin  of  time ' '  as  the  instant  at  which  the  par- 
ticle is  at  the  point  O. 

(c)  Assume  that  when  the  particle  is  at  a  distance  a  from  the 
origin  (in  positive  direction)  its  velocity  is  zero. 

To  integrate  (1),  multiply  the  first  member  by  v,  and  the  last  by 
its  equal  dx/dt;  there  results,  v(dv/dt)  =  — kx(dx\df),  or 

%>dv  =  — kxdx. 


MOTION    IN    A   STRAIGHT    LINE:    APPLICATIONS.  1 95 

Multiplying  through  by  2  and  integrating, 

v2  =  —kx2  +  C. 

Applying  condition  (V),  we  find  C  =  ka2 ;   hence  the  last  equation 
may  be  written 

vl  =  k(al—  x%        .         .         .         .     (2) 

which  gives  the  velocity  in  any  position. 

To  determine  a  relation  between  x  and  t,  a  second  integration  is 
required.      Equation  (2)  may  be  written 

(dx/dt)2  =  k{al  —x% 

or  k*dt  =  dxjVd'—x2. 

■  Integrating,  k^t  =  sm~\xla)  +  C. 

From  condition  (b)  it  follows  that  when  /  =  o,  x  =  0,  and   there- 
fore C  —  —  sin-1  (o)  =  o.      Hence  the  equation  becomes 

k%t  --—  sin_1(.r/tf), 
or  x  =  a  sin  (kl/2t) (3) 

Equation  (3)  gives  the  value  of  x  at  any  time. 

[In  determining  the  value  of  C ,  it  was  assumed  that  the  angle 
whose  sine  is  0  is  o.  But  this  is  only  one  of  many  allowable  values. 
The  general  value  would  be  sin-1  (o)  =  mr>  where  n  is  any  integer. 
Using  this  value  we  have 

whence  sin_1  (^  =  k%t  +  **' 

x\a  =  sin  (Jfat  -j-  nif)  =  sin  (k&£)  cos  ntr  -f  cos  (Jfct)  sin  nir. 

But  sin  nir  =  o,  and  cos  mr  =  ±  1  ;  the  sign  depending  on  whether 
n  is  even  or  odd.     Hence 

x  =  ±  a  sin  (kfit). 
The  double  sign  shows  that  the  assumed  conditions  are  not  sufficient 
to  fully  determine  the  motion  ;  it  is  left  uncertain  whether  the  motion 
has  the  positive  or  the  negative  direction  when  t  =  o.  The  former 
supposition  will  be  adopted,  and  the  plus  sign  used,  as  in  equation 
(3)-] 

A  discussion  of  equation  (3)  shows  that  x  is  periodic;  that  is,  in 
successive  equal  intervals  it  passes  repeatedly  through  the  same  series 
of  values.  The  greatest  value  of  x  is  a  ;  and  it  is  seen  that  x  —  a  as 
often  as  sin  (k&f)  =  1  ;  that  is,  when 

kfit  =  73-/2,     57r/2,     97J-/2,     etc. ; 

or  when  t  =  7r/2ky*,     5^/2^,     971-/2^,     etc. 


196  THEORETICAL    MECHANICS. 

The  interval  between  any  two  successive  values  of  t  in  this  series  is 
27r//£^  ;  which  is  the  interval  required  for  the  particle  to  return  to 
the  point  at  which  it  is  at  rest  after  once  leaving  it. 

Again,  the  greatest  negative  value  of  x  is  — a ;  which   occurs 
when  sin  (#*/)  =  —  1  ;  that  is,  when 

kV*t  =   37T/2,        77T/2,         II7T/2,        etC.  | 

or  when  /  =  371-/2$*,      777-/2$*,      1171-/2$*,     etc. 

The  interval  between  two  successive  values  of  /  in  this  series  is  27r/$/2. 
It  is  also  seen  that  the  interval  between  two  instants  at  which 
x  =  a  and  x  =  — a  is  7r/$/2.      Again,  the  interval  between  the  in- 
stants when  the  particle  is  at  O  and  at 

*---    a    — *---   a    *       A  is  seen  to  be  7r/2$/2.      Hence,  if  A' 

A!  0'*~JC  -4       A      is  the  point  at  which  x  =  — a,   it  is 

FlG-  io5-  seen    that,    in    successive    intervals   of 

77/2$^  sec. ,  the  particle  passes  from  A 
to  0,  from  0  to  A',  from  A'  to  0,  and  from  O  to  A  ;  and  then  re- 
peats the  cycle. 

Motion  in   accordance  with  this  law  of  force  is  called  harmonic 
motion,  and  is  of  great  importance  in  mathematical  physics. 

230.  Repulsive  Force  Proportional  Directly  to  the  Distance. — 

Consider  next  the  case  in  which  the  force  is  repulsive, —  i.  e.,  acts 
always  away  from  the  fixed  point,  —  and  let  the  magnitude  of  the 
force  vary  in  direct  ratio  with  the  distance  from  the  center  of  force. 
Let  P'  be  the  magnitude  of  the  force  when  the  particle  is  at  the 
unit  distance  from  O ;  then  P  =  P'x,  and  the  equation  of  motion  is 

m'x  =  P'x, 
or  dvjdt  =  x  =  P'x/m  =  kx,       .         .  (1) 

where  k  is  a  positive  constant,  and  means  the  magnitude  of  the  repul- 
sive force  per  unit  mass  when  the  particle  is  at  unit  distance  from  O. 
Equation  (1)  is  identical  with  the  differential  equation  for  the 
motion  when  the  force  is  attractive  (equation  (1),  Art.  229),  except 
that  — k  takes  the  place  of  k.  The  integration  may  therefore  be  car- 
ried out  in  the  same  manner ;  and  if  the  same  initial  conditions  are 
assumed,  the  final  result  will  take  the  same  form,  with  the  substitution 
of  — k  for  k.  The  position  would  be  given  by  the  equation  corres- 
ponding to  (3)  of  Art.  229: 

x  =  a  sin  (jv — k). 


MOTION    IN   A   STRAIGHT    LINE:    APPLICATIONS.  1 97 

In  order  to  avoid  this  imaginary  form,  the  integration  may  be 
effected  in  a  different  manner.  It  will  be  found  necessary  to  change 
the  initial  conditions  as  the  solution  proceeds. 

Integrating  as  in  Art.  229, 

(dx/dtj2  =  k(xl  —  d),  .         .         .     (2) 

the  constant  being  determined  from  condition  (V)  assumed  in  the  pre- 
ceding case,  that  when  x  =  a,  v  =  o.     Equation  (2)  may  be  written 


k*dt  =  dx/Vx2  —  d\ 

Integrating,  , 

k*t  =  log  (x  +  V x2  —  d)  +  C. 

If,  as  in  the  preceding  case,  it  be  assumed  that  x  =  0  when  t  =  0 
(condition  (Jj)  of  Art.  229),  the  value  of  C  is  imaginary.  This  shows 
that  such  a  condition  is  inconsistent  with  the  condition  already  as- 
sumed ;  if  the  particle  is  at  rest  when  x  =  a,  it  can  never  pass  through 
the  origin.  This  is  of  course  obvious  in  the  case  of  a  repulsive  force  ; 
it  is,  in  fact,  evident  that  x  can  never  be  less  than  a.  Therefore, 
instead  of  assuming  condition  (b)  of  Art.  229,  let  /  be  reckoned  from 
the  instant  when  x  =  a.     This  gives  C  =  — log  a,  and  therefore 

k*t  =  log  [O  -f-  Vx*  —  d)la\ 


or  x  -f  Vx2  —  a?  ==  04**'. 

Solving  for  x}  x  t« 

x  =  \a(e**  +  e~kt).  .         .         .     (3) 

Here  e  denotes  the  base  of  the  natural  system  of  logarithms,  its  value 
being  2. 7 18-}-. 

Equation  (3)  shows  that  the  motion  is  not  periodic,  as  in  the  pre- 
ceding case.  If  t  =  0,  x  =  a ;  and  this  is  the  least  value  x  can 
have.  Again,  if  /  be  given  equal  positive  and  negative  values,  the 
corresponding  values  of  x  are  equal.  Hence  the  motion  after  the 
instant  /  =  o  is  exactly  the  reverse  of  the  motion  before  that  instant. 
As  /  increases,  x  increases  ;  and  the  particle  never  returns  after  it  be- 
gins to  recede  from  its  nearest  position  to  O. 

Examples. 

1.  Let  the  force  be  attractive,  its  magnitude  at  1  ft.  from  the  center 
of  force  being  4  poundals  per  pound  of  mass  of  the  attracted  particle  ; 
and  let  the  particle  be  at  rest  at  a  certain  instant  at  10  ft.  from  the 
center.      Determine  the  position  and  velocity  in  terms  of  /. 


I98  THEORETICAL    MECHANICS. 

2.  In  the  same  case,  what  are  the  position  and  velocity  5  sec. 
after  the  particle  is  at  rest? 

Ans.  x  =  — 8.39  ft. ;  v  =  -(-10.89  ft.-per-sec. 

3.  In  the  same  case,  with  what  velocity  does  the  particle  pass  the 
center  of  force?  Ans.  20  ft.-per-sec. 

4.  How  often  does  the  particle  return  to  the  starting  point  ? 

Ans.  At  intervals  of  3. 14 16  sec. 

5.  Solve  example^  1  and  2,  assuming  the  force  to  be  repulsive, 
the  remaining  data  being  the  same  as  before. 

Ans.  When  t  =  5,  x  =  no,  100  ft.,  v  =  220,200  ft.-per-sec. 

6.  Solve  the  problem  of  the  motion  of  a  particle  under  a  force 
varying  directly  as  the  distance  from  a  fixed  point  in  the  line  of  mo- 
tion and  directed  away  from  that  point,  taking  all  data  as  in  the  above 
general  solution  with  the  following  exceptions  :  Instead  of  the  con- 
dition v  =  o  when  x  —  a,  assume  that  v  =  v0  when  x  —  o. 

7.  Solve  with  data  as  in  Ex.  5,  except  that  the  velocity  is  to  be 
10  ft.-per-sec.  when  the  particle  passes  the  center  of  force. 

8.  Let  the  force  be  attractive,  its  magnitude  at  1  met.  from  the 
center  of  force  being  1,000  dynes  per  gram  of  mass  of  the  attracted 
particle.  Let  the  particle  start  from  rest  at  150  cm.  from  the  center. 
Determine  the  position  and  velocity  at  any  time. 

9.  With  data  as  in  Ex.  8,  determine  the  velocity  when  the  par- 
ticle is  100  cm.  from  the  center  of  force,  and  determine  the  position 
and  velocity  10  sec.  later. 

Ans.  When  x  =  100  cm.,  v  —  354  cm.-per-sec 

10.  Assume  data  as  in  Ex.  8,  except  that  the  force  is  repulsive. 
Determine  the  position  and  velocity  10  sec.  after  the  particle  is  at  rest. 

231.  Force  Varying  Inversely  as  the  Square  of  the  Distance. — 
Let  a  particle  be  attracted  toward  a  fixed  point  with  a  force  whose 
magnitude  varies  inversely  as  the  square  of  the  distance  from  that 
point.  Let  P'  denote  the  magnitude  of  the  force  when  the  particle 
is  at  unit  distance  from  the  center  of  attraction  ;  then  the  equation  of 

motion  is  ••  ™ ,  2 

mx  =  — F  jx\ 

or  x  =  dxjdt  = — £/x2,  .         .         .     (1) 

in  which  k  (=  P'jm)  is  the  magnitude  of  the  attractive  force  per 
unit  mass  at  unit  distance  from  the  origin.  * 

*It  is  to  be  noticed  that  equation  (1)  applies  only  when  the  particle  is  on 
the  positive  side  of  the  origin.  The  equation  makes  the  acceleration  negative 
for  all  values  of  x,  while  in  fact  the  force  (and  therefore  the  acceleration)  is 
positive  when  x  is  negative.  Therefore  the  equation  must  be  used  for  positive 
values  of  x  only.     For  negative  values  of  x  the  equation  must  be  x  =  k/x2. 


MOTION    IN    A    STRAIGHT    LINE:    APPLICATIONS.  1 99 

Let  equation  (i)  be  integrated  subject  to  the  condition  that  the 
particle  is  at  rest  when  at  a  distance  a  from  the  origin,  and  that  /  is 
reckoned  from  that  instant.  The  first  integration  can  be  performed 
by  the  method  described  in  Art.  228. 

Multiplying  through  by  dx  and  integrating, 

\x>  =  k\x-  +  C. 

The  initial  conditions  make  the  constant  equal  to  — kja,  hence 

x2  =  (dxjdff  =  2k{\\x  —  ijd).  .         .     (2) 

Equation  (2)  gives  the  velocity  when  the  position  is  known.     To  find 
the  relation  between  x  and  /,  we  have 


dtV  2kla  =  dxVx/(a  —  x). 
Integrating 


tv  2k\a  b=3  — 1   ax  —  x'1  +  \&  sin-1  \[ix —  d)ld\  -j-  C . 
If  t  =  0  when  x  =  a  (as  above  assumed),  C  =  — -\a  sin-1  ( 1) ;  hence 


tV  2kja  =  —V  ax  —x'1  -+-  \a  sin_1[(2^ —  a)/a]  —\a  sin_1(i), 


or  tv  2k ja  =  — V  ax — x'1  -f  \a  cos-1  [(2^ — d)ja\       .       (3) 

232.  Motion  Under  the  Attraction  of  the  Earth. —  If  the  earth 
were  a  sphere  of  uniform  density  throughout,  or  a  sphere  in  which 
the  density  had  the  same  value  at  all  points  equally  distant  from  the 
center,  its  attraction  upon  any  body  outside  its  surface  would  vary 
inversely  as  the  square  of  the  distance  from  the  center.  (Art.  183.)  In 
the  actual  case  the  attraction  is  very  nearly  expressed  by  the  law  stated. 
Hence  equations  (1),  (2)  and  (3)  of  the  preceding  Article  may  be 
applied  with  small  error  to  the  motion  of  a  body  falling  vertically 
toward  the  earth  from  a  great  height,  supposing  gravity  to  be  the 
only  force  acting.  In  Art.  227  the  motion  of  a  falling  body  was  dis- 
cussed on  the  supposition  that  the  attraction  of  the  earth  upon  it  is 
constant.  In  the  present  case  the  range  of  motion  is  supposed  to  be 
so  great  that  a  more  accurate  treatment  is  desirable. 

In  order  to  apply  the  results  of  Art.  231  to  the  present  problem, 
it  is  necessary  to  determine  the  value  of  the  constant  k.  If  R  denotes 
the  radius  of  the  earth,  we  know  that  when  x  =  R  the  acceleration 
is  — g.      Hence  from  equation  (1)  we  have 

-g=~k/R\    or    k=gR\ 


200  THEORETICAL    MECHANICS. 

If  g  is  in  feet-per-second-per-second,  R  must  be  in  feet,  and  k  means 
the  attraction  in  poundals  on  a  pound-mass  if  placed  one  foot  from 
the  earth's  center  (supposing  the  same  law  of  variation  of  the  earth's 
attractive  force  to  hold  within  the  surface  as  without).  The  value  of 
R  may  be  taken  as  20,900,000  feet. 

Examples. 

1.  A  body  starts  from  rest  at  a  height  above  the  surface  equal 
to  the  earth's  radius.  Compute  the  velocity  when  the  surface  is 
reached. 

Putting  k  =  gR2,  the  velocity  is  given  by  equation  (2)  (Art.  231  J, 
which  may  be  written 

v2  =  2gR\\\x  —  i/a). 
In  this  example  a  =  2R  ;  hence  for  x  =  R  we  have 
v2  =  2gR\i/R  —  1/2R)  =gR. 

In  feet-per-second, 

v  =  V 32.2  X  20,900,000  =  25,900. 

2.  What  is  the  greatest  velocity  a  body  could  acquire  in  falling 
from  rest  to  the  earth's  surface ?     [Put  a  =  do,  x  =  R.~\ 

3.  With  what  velocity  must  a  body  be  projected  upward  at  the 
earth's  surface  in  order  that  it  may  never  return? 

Ans.  A  velocity  not  less  than  6.95  miles-per-sec. 

4.  Deduce  a  formula  for  the  velocity  acquired  by  a  body  in  falling 
to  the  surface  from  a  height  h. 

Putting  a  =  R  +  h  and  x  —  R,  the  above  general  formula  for 
v2  becomes  ^  =  2gk[RI{R  +  h)\ 

If  h  is  small  compared  with  R,  we  may  put  as  a  first  approxima- 
tion ^/(^  +  fy p-  J  ;     ...      7r>  =  2g/L 

This  is  identical  with  the  formula  which  would  apply  if  the  attraction 
were  constant.      See  Ex.  1,  Art.  227. 

In  using  the  accurate  formula,  if  hIR  is  small  we  may  write 

R/(R  +  h)  =  (1  +  /1/R)-1 
and  therefore 

v2  =  2g/i[i  -wr)  +  WRy-(A/Ry+   .    .    .    ]. 

By  taking  any  number  of  terms  of  this  series  an  approximate  result 
may    be  obtained  which  is  correct  to  any  desired  degree. 

5.  Let  a  body  fall  to  the  surface  from  a  height  of  5,000  ft. 
Compute  the  velocity  acquired,  using  first  the  approximate  formula 
and  second  the  accurate  formula.      (Take  g  =  32. 2.) 

6.  Determine  the  value  of  fc,  using  C.  G.  S.  units. 


MOTION    IN    A    STRAIGHT    LINE  I    APPLICATIONS.  201 

§  4.   Miscellaneous  Problems. 

233.  Constant  Force  Combined  With  Central  Force. —  Let  a 

body  be  acted  upon  simultaneously  by  a  constant  force  and  a  force 
directed  toward  a  fixed  point.  Then  if  Pl  and  Pt  are  the  values  of 
the  two  forces  at  any  instant,  the  general  equation  becomes 

**='/>+ i>„  •     (0 

and  we  must  put  for  Px  and  P2  their  values  as  functions  of  x  and  /. 
If  the  origin  is  taken  at  the  center  of  force,  and  the  central  force 
is  a  function  of  the  distance  from  the  center,  we  have 

Px  =  constant;     P2  =/(*). 

The  only  case  that  will  be  discussed  is  the  following : 

Body  suspended  by  elastic  string.  —  A  body  is  suspended  from  a 
fixed  point  by  an  elastic  string,  and  is  acted  upon  by  no  force  except 
its  weight  and  the  supporting  force  exerted  by  the  string.  To  de- 
termine the  motion. 

We  must  here  make  use  of  the  property  that  a  stretched  elastic 
string  exerts  a  resisting  force  proportional  directly  to  the 
amount  of  stretching.     This  law  applies  not  only  to  elastic    My. 
strings  but  to  a  bar  or  rod  of  any  elastic  material.      The 
law  is  established  by  experiment,  and  for  many  substances 
is  nearly,  though  perhaps  in  no  case  exactly,  true. 

Let  MO  (Fig.  106)  represent  the  unstretched  string, 
M  being  the  point  of  support.      Let  /  denote  the  length     0 
MO,   and  x  the  amount  of  stretching  at  the  time  / ;  at 
which  instant  the  end  of  the  string,  originally  at  0,  is  at 
A.     The  upward   force  exerted    by  the  string  upon  the 
particle  is  then  proportional  to  x,  and  we  may  put  P2  = 
—  ex  in  equation  (1),  c  being  the  force  necessary  to  pro-     ^ 
duce  an  elongation  of  one  unit  of  length.     Since  the  force    pIG   Io6 
producing  a  given  elongation  is  greater  in  proportion  as 
the  unstretched  length  is  less,  c  is  inversely  proportional  to  /,  and 

we  may  write 

.  P%  ~  —(e/l)xy 

e  being  a  constant. 

For  Px  may  be  written  mg,  the  weight  of  the  body  ;  hence  equa- 
tion (1)  becomes 

m(d'2x/dt2)  =  mg  —  (e/l)x.     .         .         .     (2) 


202  THEORETICAL    MECHANICS. 

This  equation  can  be  integrated  directly,  but  the  process  is  simpler 
if  the  origin  of  coordinates  is  changed.  Let  0'  (Fig.  106)  be  the 
position  of  the  end  of  the  string  when  the  two  forces  just  balance 
each  other  ;  the  distance  of  this  point  from  O  is  the  value  of  x  found 
by  putting  Px  -\-  P2  =  o,  that  is 

mg  —  exjl  =  o,     or    x  =  mgl/e. 
Introducing  a  new  variable  *,  such  that 

x  =  z  +  mglje, 
equation  (2)  becomes 

d2z/dt2  =  —(e/ml)z. 

The  variable  z  evidently  means  the  abscissa  of  the  particle  measured 
from  O'.  The  last  equation  is  identical  in  form  with  equation  (1)  of 
Art.  229,  and  the  solution  there  given  is  here  applicable.  It  is  thus 
seen  that  in  the  case  now  under  discussion  the  motion  is  the  same  as 
if  the  particle  were  acted  upon  by  a  single  force  directed  toward  the 
point  O'  and  varying  directly  as  the  distance  from  that  point. 

Examples. 

1.  If  a  force  equal  to  10  pounds- weight  would  change  the  length 
of  the  string  from  5  ft.  (its  natural  length)  to  6  ft,  and  if  the  mass 
of  the  body  is  5  lbs. ,  write  the  equation  of  motion. 

A ns.   d'2x/dt2  —  (1  —  2X)g,  or  d2zjdt2  =  —2gz. 

2.  In  the  same  case,  let  the  body  be  at  rest  when  the  string  is 
unstretched  ;  determine  the  motion  completely. 

Ans.  z  =  — \  cos  (ty  2g). 

3.  Can  the  initial  conditions  be  such  that  the  equation  deduced 
above  does  not  apply  throughout  the  whole  of  the  motion?  If  so, 
how  could  such  a  case  be  treated  ? 

[The  value  of  P.2 ,  the  upward  pull  of  the  string  on  the  body,  be- 
comes zero  when  the  string  shortens  to  its  natural  length,  the  body 
being  then  at  O  (Fig.  106).  If  the  body  rises  above  this  point,  Pt 
is  zero,  and  so  long  as  this  is  the  case  the  body  moves  under  the 
single  force  of  gravity. 

If  the  string  be  replaced  by  a  rigid  bar  of  elastic  material  (such 
as  steel  or  wrought  iron  when  stretched  within  the  ' '  limit  of  elas- 
ticity"), the  force  P2  will  follow  the  same  law  when  the  bar  is  short- 
ened as  when  it  is  lengthened;  that  is,  — exjl  is  the  value  of  P2 
throughout  the  whole  motion,*  even  if  the  initial  conditions  are  such 
that  the  length  of  the  bar  becomes  less  than  the  natural  length.] 

*  This  assumes  that  the  "  modulus  of  elasticity  "  has  the  same  value  in 
compression  as  in  tension. 


MOTION    IN    A    STRAIGHT    LINE  :    APPLICATIONS.  203 

4.  With  the  data  of  Ex.  1,  let  the  body  have  a  velocity  of  10  ft.- 
per-sec.  at  the  instant  when  the  string  is  unstretched.  Determine 
the  motion  completely. 

234.  Motion  in  a  Resisting  Medium — If  a  body  be  moving 
through  the  air,  or  through  any  fluid,  the  fluid  exerts  upon  it  forces 
which  depend  upon  the  velocity.  The  effect  of  these  forces  is  to 
retard  the  motion  of  the  body ;  in  other  words,  the  resistance  of 
the  medium  is  equivalent  to  a  force  whose  direction  is  always  op- 
posite to  that  in  which  the  body  is  moving.  Hence  in  the  general 
equation  of  rectilinear  motion,  such  a  resistance  will  enter  as  a  force 
whose  magnitude  is  some  function  of  the  velocity,  and  whose  direc- 
tion is  opposite  to  that  of  the  velocity  ;  that  is,  for  such  a  force, 

care  being  taken  that  the  correct  sign  is  used 

Examples. 

1.  Write  the  equation  of  motion  for  a  body  moving  vertically 
under  gravity  and  the  resistance  of  the  air,  the  latter  being  assumed 
to  vary  directly  as  the  velocity.  Explain  the  meanings  of  any  con- 
stants entering  the  equation.      Integrate  the  equation  completely. 

2.  In  the  same  case,  assume  the  resistance  of  the  air  to  vary  as 
the  square  of  the  velocity.  Can  a  single  equation  be  written  for  the 
whole  motion  in  this  case,  if  the  body  has  initially  an  upward  velocity? 
Integrate  the  equation  once,  thus  determining  the  relation  between 
velocity  and  time. 

235.  Motion  of  Connected  Particles.— If  two  particles  are  con- 
nected by  a  string  which  is  kept  tight,  each  is  acted  upon  by  a  force 
due  to  the  string.  If  the  string  is  without  weight  and  is  not  in  con- 
tact with  any  body  except  the  two  particles  mentioned,  the  tension 
sustained  by  it  has  the  same  value  at  every  cross-section.  Hence 
the  string  exerts  equal  and  opposite  forces  upon  the  particles. 

If  the  string  passes  around  smooth  pegs  or  pulleys,  the  tension  is 
still  uniform  throughout  its  length,  and  the  forces  exerted  upon  the 
particles  by  the  string  are  equal  though  they  may  not  be  opposite. 

If  the  equation  of  motion  be  written  for  each  particle,  each  equa- 
tion will  contain  the  force  due  to  the  tension  of  the  string.  By  com- 
bining the  two  equations  this  unknown  force  may  be  eliminated. 
This  will  be  illustrated  by  a  particular  case. 

Let  two  particles  whose  masses  are  my  and  m%  be  connected  by  a 


204  THEORETICAL    MECHANICS. 

string  which  passes  over  a  smooth  pulley.  Let  the  initial  conditions 
be  such  that  each  particle  describes  a  vertical  line,  one  rising  while 
the  other  falls.  Assume  the  string  to  be  without  weight,  perfectly 
flexible  and  inextensible.  Assume  also  that  the 
mass  of  the  pulley  is  so  small  as  to  be  negligible, 
and  that  it  can  revolve  without  frictional  resistance. 
The  arrangement  is  shown  in  Fig.  107. 

Let  mx  be  greater  than  m2 ,  and  assume  the  par- 
ticles to  be  at  rest  at  a  certain  instant.      Then  mx 
will  fall  and  m2  will  rise.      Take  the  position  of  rest 
as  origin  for  each  particle,  and  let  x  denote  the  dis- 
'~\tti        !~]m     tance  °f  eacn  fr°m  its  origin  at  the  time  t.      If  T  = 
Fig.  107.  tension  in  string,  the  resultant  force  acting  upon  mx 

is  mxg  —  T  downward,  and  the  resultant  force  act- 
ing upon  m2  is  T  —  m2g  upward.  The  acceleration  of  mx  is  x 
downward,  and  that  of  m2  is  x  upward.  The  equations  of  motion 
for  the  two  particles  are  therefore 

mxg  —  T  =  mxx;    .         .         .         .     (1) 
T  —  m2g  --=  m2x.     .  .         .  (2) 

The  unknown  quantity  T  may  be  eliminated  by  adding  the  two 
equations.     The  result  is 

(mx  —  m2)g  =  {mx  -f  m2)x.  .  .  .  (3) 
This  equation  is  the  same  as  would  apply  to  the  motion  of  a 
particle  of  mass  mx  -\-  mt  acted  upon  by  a  force  {inx  —  m.^g ;  i.  e. , 
by  a  force  equal  to  the  difference  between  the  weights  of  the  particles. 
It  is  as  if  the  combined  mass  of  the  two  particles  were  being  pulled 
in  opposite  directions  by  forces  mxg  and  m.,g. 

Evidently  equation  (3)  may  be  treated  by  the  methods  employed 
in  Arts.  226  and  227.  The  equations  derived  from  the  motion  of  a 
falling  body  may  be  made  applicable  to  the  present  case  by  substitut- 
ing [(**,  —  m2)/(mx  +  m.2)]g  for  g. 

Value  of  the  tension. —  By  eliminating  x  between  any  two  of 
equations  (1),  (2)  and  (3),  the  value  of  T  is  found.      It  is 

T  =  2[mxmJ(mx  -f-  m2)]g.      .         .         .     (4) 

Examples. 

1.  If  the  two  masses  are  4  lbs.  and  4. 1  lbs.,  determine  the  accel- 
eration, the  tension,  the  velocity  acquired  after  0.5  sec,  and  the 
distance  fallen  through  in  0.5  sec.        Ans.  Acceleration  =  g/81. 


MOTION    IN    A    STRAIGHT    LINE  :    APPLICATIONS. 


205 


2.  Two  masses  of  2  kilogr.  each  are  suspended  from  the  ends  of 
a  string  which  passes  over  a  smooth  pulley.  The  system  being  at 
rest,  a  mass  of  10  gr.  is  added  to  one  side.  Determine  the  subse- 
quent motion. 

3.  Three  particles  are  connected  by  two  strings,  one  of  which 
passes  over  a  smooth  pulley  as  in  Fig.  108.     De- 
termine the  motion,  and  the  tensions  in  the  strings. 

Let  mx ,  m2  and  m3  denote  the  masses  of  the 
three  particles,  Tx  the  tension  in  the  string  mxm2 
(Fig.  108),  and  T2  the  tension  in  m2m.A.  Let  x  be 
the  distance  of  mz  below  a  certain  fixed  point ; 
then  x  is  also  the  distance  of  mx  above  some  fixed 
point,  and  of  m%  above  another  fixed  point.  The 
equations  of  motion  for  the  three  particles  are 
Tx  —  mxg  =  mxx  ; 
T2  —  Tx  —  m2g  =  m^x ; 


□m2    \_Jm3 


m. 


g 


T2  =  mJc, 


The  unknown  quantities  Tx  and  T2  may  be  elimi-      I      ira, 
nated  by  adding  the  three  equations.      The  result  Fig.  108. 

ls      (m3  —  mx  —  tn^)g  =  (mx  -\-  m2  -f-  m3)x. 

This  is  identical  with  the  equation  of  motion  for  a  particle  of  mass 
mx  -f-  m2  -f-  m.A  acted  upon  by  a  force  m^g  in  one  direction  and  by 
forces  mxg,  m2g  in  the  opposite  direction. 
The  values  of  Tx  and  T2  are 

Tx  =  mx(g  +  x)  =  2[mxm3/(mx  +  m2  +  m^g; 
T2  =  m,(g  —  x)  =  2[01  -j-  m2)m.J(mx  +  m2  -f  m^\g. 

4.  In  Ex.  3,  let  mXi  m2  and  m.Ai  expressed  in  kilograms,  have 
values  2,  3  and  4.      Determine  the  acceleration  and  the  tensions. 

Arts.   Tx  =  16/9  kilograms-weight,  T2  =  40/9  kilograms-weight. 

5.  Let  mx  =  4  lbs.,  m2  =  3  lbs.,  m3  =  2  lbs.     Determine   Tx, 
T2  and  the  acceleration. 

Ans.  The  acceleration  of;;/,  is  5^/9  downward. 


Miscellaneous  Examples. 

[In  the  following  examples  the  student  should  in  every  case  write 
the  differential  equation  of  motion,  even  if  it  is  not  found  possible  to 
complete  the  solution.  He  should  also  examine  what  conditions  are 
needed  for  the  determination  of  the  constants  of  integration.  If  this 
is  done,  the  problem  in  Dynamics  is  reduced  to  one  in  mathematical 
analysis.] 

1 .  Two  particles  whose  masses  are  equal  are  connected  by  an 
elastic  string.  They  are  projected  from  the  same  point  with  equal 
and  opposite  velocities.  Determine  the  motion,  assuming  no  forces 
to  act  except  those  due  to  the  tension  of  the  string. 


206  THEORETICAL    MECHANICS. 

Ans.  Let  a  be  the  natural  length  of  the  string  and  k  the  force 
necessary  to  double  its  length.  If  A  is  the  position  of  one  particle 
when  the  string  begins  to  stretch,  that  particle  describes  half  of  a 
harmonic  oscillation  (Art.  229)  about  A,  returning  to  A  after  a  time 
7Ti/(am/2k).  The  two  particles  then  approach  each  other  with 
constant  velocity. 

2.  Assume  the  conditions  as  in  Ex.  1,  except  that  the  masses  are 
unequal. 

3.  Two  particles  whose  masses  are  ml  and  m2 ,  connected  by  an 
inextensible  string,  are  suspended  from  a  fixed  point  by  an  elastic 
string  attached  to  mx .  If  the  system  is  held  at  rest  with  the  elastic 
string  at  its  natural  length  and  then  released,  determine  the  subse- 
quent motion.  Determine  also  the  tension  in  the  lower  string  in  the 
lowest  position. 

Ans.  The  required  tension  is  twice  the  weight  of  mt. 

4.  Two  particles,  each  of  mass  m,  are  connected  by  an  elastic 
string  whose  natural  length  is  /.  The  force  necessary  to  double  the 
length  of  the  string  is  F.  The  particles  repel  each  other  with  forces 
varying  directly  as  their  distance  apart ;  the  repulsive  force  being  F' 
when  the  distance  is  /.  If  both  particles  are  held  at  rest  at  a  dis- 
tance apart  equal  to  /and  are  then  released,  determine  the  subsequent 
motion. 

Ans.  \i  F  ^>  F' ,  the  motion  of  either  particle  is  given  by  the 
equation  

x  _  F  +  F'  sin  [tVi^F—  F")/ml~~\ 

l~  2(F—F') 

The  motion  is  a  harmonic  oscillation  (Art.  229)  about  a  point  distant 
F//2(F  —  F')  from  the  point  midway  between  the  particles,  the  time 
of  a  complete  oscillation  being  wy/\_2m//{F  —  F')\  If  the  origin  be 
taken  at  the  center  about  which  the  oscillation  takes  place,  the  motion 
is  given  by  equation  (3)  of  Art.  229,  if  k  =  2(F — F')jml.  If 
F'  >  Fy  the  motion  reduces  to  a  case  under  Art.  230. 

5.  Solve  with  data  as  in  Ex.  4  except  that  the  masses  are  un- 
equal. 

6.  In  Ex.  4  let  the  mass  of  each  particle  be  40  gr. ;  let  /  =  20 
cm.,  F  =  50 grams-weight,  F'  =  30 grams-weight.  Determine  the 
period  of  a  complete  oscillation,  and  the  range  of  motion. 

Ans.  Time  of  oscillation  =  0.9  sec. 

7.  In  Fig.  108  let  the  string  mlm2  be  elastic,  its  unstretched 
length  being  /.  Let  the  system  be  initially  at  rest  with  mlm2  =  /. 
Determine  the  subsequent  motion. 

8.  Two  particles,  each  of  mass  m,  repel  each  other  with  forces 
varying  inversely  as  the  square  of  their  distance  apart.  When  the 
distance  is  /  the  force  is  F.     They  are  held  at  rest  in  a  vertical  line 


MOTION    IN    A   STRAIGHT    LINE  :    APPLICATIONS.  207 

at  a  distance  apart  /',  and  are  then  released.  Determine  their  subse- 
quent motion,  assuming  the  only  forces  acting  on  the  particles  to  be 
gravity  and  their  mutual  repulsion. 

Ans.  If  r  is  the  distance  of  the  particles  apart  at  any  time,  their 
velocity  of  separation  drjdt  is  given  by  the  equation 

{drfdff  =  UFP/m)(i//f  —  i/r). 
This   equation   can   be   integrated,  giving  /asa  function  of  r.     The 
center  of  gravity  of  the  two  particles  falls  a  distance  \gt2  in  /sec. 
These  two  relations  determine  the  motion. 

9.  A  particle  of  mass  m  is  attached  to  one  end  of  an  elastic 
string  whose  natural  length  is  /.  The  other  end  of  the  string  is  at- 
tached to  a  fixed  support.  The  particle  is  dropped  from  this  fixed 
point  of  attachment.  Determine  the  motion.  Let  the  force  neces- 
sary to  stretch  the  string  to  double  its  natural  length  be  F. 

10.  The  natural  length  of  an  elastic  string  is  /;■  under  a  pull  Fit 
stretches  to  a  length  2/.  The  ends  are  attached  to  fixed  pegs  whose 
distance  apart  is  37/2.  A  particle  of  mass  m  is  attached  to  the  string 
at  a  point  distant  4// 5  from  one  peg,  and  is  forcibly  brought  to  the 
point  midway  between  the  pegs  and  then  released.  Determine  the 
time  of  oscillation.  Ans.   fair/ 15))/ (14m// F). 

11.  A  body  is  projected  into  a  resisting  medium  which  exerts  a 
retarding  force  proportional  to  the  velocity.  If  no  other  force  acts 
upon  the  body,  determine  the  motion.  Let  m  =  mass,  V  ==  initial 
velocity,  and  let  the  force  =  F  when  the  velocity  ==  V. 

Ans.  Let;r'  denote  the  distance  the  body  would  move,  and  t'  the 
time  it  would  move,  before  coming  to  rest  against  a  constant  force  F. 
Then  the  distance  described  in  time  t  against  the  actual  force  is 
x=  ix' {1  — e~flt').  It  is  seen  that  x  approaches  the  limit  2x ', 
but  the  particle  never  comes  to  rest. 

12.  A  mass  of  5  kilogr.  moves  in  such  a  way  that  its  acceleration 
is  directly  proportional  to  its  velocity  but  has  the  opposite  direction. 
When  expressed  in  centimeter-second  units  the  velocity  and  accelera- 
tion are  numerically  equal.  In  a  certain  position  the  velocity  is  50 
c.m.-per-sec.  Determine  the  value  of  the  force  after  the  particle  has 
moved  40  cm.  from  this  position.  Ans.  50,000  dynes. 

13.  In  Ex.  12,  determine  the  value  of  the  force  1  sec.  after  the 
instant  at  which  the  velocity  is  50  c.  m.  -per-sec.  When  will  the  par- 
ticle come  to  rest,  and  how  far  will  it  move  ? 

Ans.  92,000  dynes.  The  distance  passed  over  will  approach  50 
cm.  as  a  limit,  but  the  particle  will  never  come  to  rest. 


CHAPTER   XIV. 

MOTION    IN    A    CURVED    PATH. 

§  i.   Position,  Displacement  and  Velocity. 

236.  Direction. —  When  the  path  of  a  particle  is  a  straight  line, 
its  motion  has  at  every  instant  one  of  two  opposite  directions.  These 
two  directions  may  be  fully  specified  by  signs  plus  and  minus.  But 
when  the  path  is  a  curve,  the  direction  of  the  motion  continually 
changes,  and  cannot  be  specified  so  simply. 

In  the  foregoing  analysis  of  the  motion  of  a  particle  whose  path  is 
a  straight  line,  definitions  have  been  given  of  displacement,  velocity 
and  acceleration.  These  definitions  must  now  be  enlarged.  Each 
one  of  these  quantities  is  at  every  instant  associated  with  a  definite 
direction  in  space,  and  this  direction  is  an  essential  element  in  its 
value.    The  quantities  named  are,  in  fact,  vector  quantities  (Art.  16). 

237.  Position  Given  by  Vector.—  If  a  particle  is  not  confined 
to  a  straight  line,  but  has  any  motion  in  space,  its  position  at  any  in- 
stant may  be  specified  by  means  of  a  vector  of  position. 

UP  (Fig.  109)  is  the  position  of  the  particle  at  any  instant,  and 

0  any  fixed  point  taken  as  origin  of  reference,  the  position  of  Pis 

known  if  the  vector  OP  is  known.      For,  to  know  the  vector  OP 

completely  is  to  know  (1)  the  direction  from  O  to  P  and  (2)  the 

length  of  the  line  OP.      Knowing  these,  P 

may  be  located  from  O.      OP  is  called  the 

position-vector  of  the  particle. 

As  the  particle  moves,  the  position-vec- 
tor varies  either  in  length,  or  in  direction, 
or  in  both  length  and  direction. 
pIG   IO  Practically,  in  order  to  describe  com- 

pletely the  magnitude  and  direction  of  a 
vector,  the  values  of  certain  angles  and  distances,  measured  from 
definite  lines,  planes  or  points  regarded  as  fixed,  must  be  given. 
For  the  present  purpose  it  is  not  necessary  to  consider  how  the 
value  of  a  vector  may  best  be  specified  in  practice. 

238.  Displacement. —  If  the  particle  moves  from  A  to  B  during 
any  interval  of  time,  the  vector  A  B  is  its  displacement. 


MOTION    IN    A    CURVED    PATH.  209 

This  definition  is  independent  of  the  actual  path  followed,  which 
may  be  any  line  joining  A  and  B.  If  the  path  is  the  straight  line 
AB  (Fig.  no),  the  vector  AB  is  the 
actual  displacement.  If  any  other  path 
ACB  is  followed,  the  vector  AB  is  still 
regarded  as  the  total  or  resultant  dis- 
placement. 

239.  Uniform  Motion. —  The  mo- 
tion is  said  to  be  uniform  when  the 

particle  receives  equal  displacements  in  any  equal  intervals  of  time, 
however  these  intervals  be  chosen. 

It  is  to  be  particularly  noticed  that,  since  displacement  is  a  vector 
quantity,  successive  displacements  are  not  equal  unless  they  agree  in 
direction  as  well  as  in  magnitude.  Hence,  uniform  motion  as  here 
defined  has  a  constant  direction  ;  i.  e. ,  it  is  rectilinear.  This  case  of 
motion  has  been  considered  in  Chapters  XII  and  XIII. 

240.  Variable  Motion. —  If  the  particle  receives  unequal  displace- 
ments in  equal  intervals  of  time,  the  motion  is  variable.  The  un- 
equal displacements  may  differ  in  length  only,  in  direction  only,  or 
in  both  length  and  direction. 

If  the  successive  displacements  agree  in  direction,  the  path  is  a 
straight  line,  and  the  motion  falls  under  the  case  already  treated. 
The  general  analysis  which  follows  includes  this  as  a  special  case. 

241.  Velocity. —  In  case  of  motion  not  confined  to  a  straight 
line,  the  velocity  of  a  particle  may  still  be  defined  as  its  rate  of  dis- 
placement /*  but  the  definition  must  be  interpreted  with  reference  to 
the  meaning  of  displacement  as  a  vector  quantity. 

Whatever  path  a  particle  may  describe,  it  is  at  every  instant 
moving  at  a  definite  rate  and  in  a  definite  direction.  The  velocity  is 
a  vector  quantity  whose  direction  coincides  with  that  in  which  the 
particle  is  moving,  and  whose  magnitude  measures  the  rate  of  motion. 

The  meaning  of  the  definition  of  velocity  and  the  method  of  esti- 
mating its  value  are  best  understood  by  considering  "average" 
velocity  in  case  of  the  unrestricted  motion  of  a  particle. 

242.  Average  Velocity. —  The  average  velocity  of  a  particle  for 
an  interval  of  time  during  which  it  receives  any  displacement,  is  the 

*As  in  rectilinear  motion,  Art.  190. 
14 


2IO  THEORETICAL    MECHANICS. 

velocity  of  a  particle  which,  moving  uniformly,  would  receive  the 
same  total  displacement  in  the  same  time. 

The  average  velocity  is  a  vector  quantity,  its  direction  being  that 
of  the  total  displacement  for  the  interval. 

Note  that  this  definition  is  inde- 

£^^  pendent  of  the  path.     Thus,   if  the 

particle  moves  from   A   to  B  (Fig. 

Fig.  in.  in)  along  any  path  ACB,  the  total 

displacement  is  the  vector  AB.     If  tx 

and  t2  are  the  values  of  t  corresponding  to  the  positions  A  and  B, 

the  average  velocity  is  given  by  the  expression 

(vector  AB)l(t2  —  t,), 

whatever  the  form  of  the  path  A  CB. 

243.  Approximate  Value  of  Velocity  at  an  Instant. —  An  ap- 
proximate value  of  the  velocity  of  a  particle  at  any  instant  may  be 
found  by  determining  the  average  velocity  for  a  very  short  time. 
Thus,  let  A  (Fig.  in)  be  the  position  of  the  particle  at  the  time  t} 
and  B  its  position  after  a  short  interval  At  ;  then 

(vector  AB)/At 

is  the  average  velocity  for  the  interval  A/,  and  is  an  approximate 
value  of  the  velocity  in  the  position  A.  The  approximation  is  closer 
the  shorter  the  interval  At. 

244.  Exact  Value  of  Velocity  at  an  Instant. —  The  true  value 
of  the  velocity  at  the  instant  /,  when  the  particle  is  at  A,  is  the  limit 
approached  by  the  above  approximate  value  as  At  approaches  zero. 
That  is 

velocity  at  A  =  limit  [(vector  AB)'At]. 

This  limit  is  a  vector  quantity. 

(1)  Its  direction  is  that  of  the  tangent  to  the  path  at  A,  since  the 
chord  AB  approaches  the  tangent  as  B  approaches  A. 

(2)  Its  magnitude  is  ds/dt,  if  s  denotes  the  length  of  the  path 
measured  from  some  fixed  point  to  the  position  of  the  particle.  For 
as  the  point  B  approaches  A,  the  chord  AB  and  the  arc  AB  ap- 
proach a  ratio  of  equality,  so  that 

limit  [(chord  AB)/At]  =  limit  [(arc  AB)/At]  =  ds/dt. 

The  magnitude  of  the  velocity,  considered  without  reference  to 


MOTION    IN    A    CURVED    PATH.  211 

direction,  is  called  the  speed.     If  this  is  denoted  by  v}  its  value  is  al- 
ways given  by  the  equation 

v  =  dsjdt. 

245.  Graphical  Representation  of  Velocity ;  Hodograph. —  Let 
AB  (Fig.  112)  be  the  path  of  a  par- 
ticle, described  in  any  manner.    From  jp 
some  point  0'  draw  O'A'  to  repre-    '/^"~"~~  -^^^ 
sent  in  magnitude  and  direction  the  J^r 
velocity  of  the  particle  in  the  position                       ^^^^^  V 
At  and  O'B'  to  represent  the  velocity    n,^^^^^  \ 
in  the  position  B.     Also,  for  every        ^""""^J                        r\P' 
intermediate  position  of  the  particle,                     ^*\^^  \ 
as  P,  draw  a  vector  (9'P'  represent-                               ^^"\^^    1 
ing  the  velocity  of  the  particle  when  no 
in  that  position.     Through  the  ex-                      Fig.  II2- 
tremities  of  all  such  vectors  draw  a 

curve.     This  is  called  the  curve  of  velocities  or  hodograph  of  the 
motion. 

Examples. 

i  .  A  particle  describes  a  circle  of  2  ft.  radius  with  uniform  speed, 
the  whole  circumference  being  described  in  half  a  second.  Required 
(a)  the  speed  ;  (Jj)  the  values  of  the  average  velocity  for  1/10  sec. 
and  for  1/8  sec. ;  (V)  the  values  (both  direction  and  magnitude)  of 
the  instantaneous  velocity  at  two  instants  1/10  sec.  apart. 

2.  A  particle  describes  a  circle  of  radius  r  with  uniform  speed  v. 
Determine  the  magnitude  and  direction  of  the  average  velocity  for  an 
interval  t.  Ans.   Its  magnitude  is  (2r/t)  sin  (vt/2r). 

3.  A  particle  describes  a  circle  of  radius  r  in  such  a  way  that 
s  =  af\  s  being  the  length  of  the  path  described  in  time  t,  and  a 
being  a  constant.  Required  (a)  the  speed  at  any  time ;  rb)  the 
average  velocity  during  the  interval  from  tx  to  /2 . 

4.  In  Ex.  3,  let  r  =  60  c.  m. ,  and  let  the  length  of  arc  described 
during  the  first  second  be  1 20  c.  m.  Required  (a)  the  speed  at  the 
end  of  2  sec. ;  (b)  the  average  velocity  during  an  interval  of  o.  1  sec. 
after  /=  2.     Ans.  {a)  48oc.m.-per-sec.    ifi)  478.5  cm. -per-sec. 

5.  Draw  the  hodograph  for  the  motion  described  in  Ex.  1. 

6.  Draw  the  hodograph  for  the  motion  described  in  Ex.  4. 

7.  A  particle  describes  any  curved  path  with  uniform  speed  v. 
What  is  the  form  of  the  hodograph  ? 


212  THEORETICAL    MECHANICS. 

§  2.    Velocity -Increment  and  Acceleration. 

246.  Increment  of  Velocity.  —  Let  the  values  of  the  velocity  of 
a  moving  particle  at  two  instants  tx  and  tt  be  represented  by  the  vec- 
tors O'A  and  O'B'  respectively  (Fig.  113).  Then  the  vector  A'B' 
is  the  velocity-increment  for  the  interval  from  tx  to  t.2.  For  A'B' 
is  the  vector  which  must  be  added  to  O'A'  to  produce  O'B'.  (See 
Art.  22.) 

It  will  be  seen  that  velocity-increment  as  thus  defined  is  not  the 
amount  by  which  the  speed  changes.     The  increment  of  the  speed  is 

length  O'B'  —  length  O'A'  =  v2  —  vly 

if  vx  and  v%  are  the  initial  and  final  values  of  the  speed.  It  is  only 
when  the  initial  and  final  velocities  are  parallel  that  the  velocity- 
increment  \s  equal  in  magnitude  to  v2  —  vx. 

247.  Acceleration. —  If  a  particle  is  moving  in  a  curved  path, 
the  acceleration  may  be  defined  as  the  rate  of  change  of  the  velocity \ 
just  as  in  the  case  of  rectilinear  motion  (Art.  203).  But  since 
"change  of  velocity "  (or  velocity-increment)  is  a  vector  quantity, 
acceleration  is  also  a  vector  quantity ;  and  in  computing  its  value 

both  direction  and  magnitude  must  be 
considered. 

In    this    definition    of   acceleration, 
"rate  of  change  of  velocity"  must  be 
understood  to  mean  ' '  increment  of  ve- 
locity per  unit  time." 
Fig.  113.  Consider  first  the  case  in  which  the 

velocity  is  varying  uniformly  ;  by  this  is 
meant  that  the  increments  of  velocity  during  any  different  intervals 
of  time  have  the  same  direction  and  are  proportional  to  the  intervals. 
Thus,  let  the  vector  A'B'  (Fig.  113)  represent  the  velocity-increment 
during  an  interval  of  time  At,  and  let  the  velocity-increments  during 
any  partial  intervals  into  which  At  may  be  divided  have  the  direction 
A'B'  and  be  proportional  in  magnitude  to  the  'partial  intervals. 
Then  the  acceleration  as  above  defined  has  a  constant  value  through- 
out the  time  At,  that  value  being 

(vector  A'B')/At. 
When  the  velocity  is  not  varying  in  this  uniform  manner,  the 


MOTION    IN    A    CURVED    PATH.  213 

acceleration  is  a  variable  vector  quantity,  having  a  definite  mag- 
nitude and  direction  at  any  instant.  The  exact  meaning  of  the 
definition  of  acceleration  in  this  case,  and  the  method  of  computing 
its  value,  may  be  best  understood  by  a  consideration  of  "average 
acceleration." 

248.  Average  Acceleration.  —  The  average  acceleration  of  a  par- 
ticle   for   an    interval    of  time    during 

which  the  velocity  varies  in  any  man- 
ner is  an  acceleration  which,  if  con- 
stant in  magnitude  and  direction, 
would  result  in  the  same  velocity- 
increment  in  the  same  time.*  The 
value  of  the  average  acceleration  is 
found  by  dividing  the  velocity-incre- 
ment received  during  the  whole  in- 
terval by  the  duration  of  the  interval. 

\{  A B'  (Fig.  114)  represents  the  velocity-increment  for  the  in- 
terval from  ty  to  t2 ,  the  expression 

(vector  A'B')/(t2  —  tx)     or     (vector  A'B')/At 

gives  the  average  acceleration  for  that  interval. 

This  expression  cannot  be  reduced,  as  in  the  case  of  rectilinear 
motion  (Art.  208),  to  the  form 

in  which  vx  and  v2  denote  the  initial  and  final  values  of  the  magnitude 
of  the  velocity  ;  for  it  is  only  when  the  two  velocities  are  parallel  that 
v2  —  vx  gives  the  velocity-increment.  \ 

249.  Approximate  Value  of  Acceleration  at  an  Instant. —  The 

average  acceleration  for  a  short  interval  immediately  following  any 
instant  is  an  approximate  value  of  the  true  acceleration  tf/that  instant. 
Thus,  if  the  velocity  at  the  given   instant  is  represented  by  O'A' 

*  Compare  Art.  208. 

t  If,  however,  i\  and  v2  be  regarded  as  vector  symbols  representing  the 
initial  and  final  values  of  the  velocity  in  direction  as  well  as  in  magnitude,  the 
vector  v2  —  vx  is  the  value  of  the  velocity-increment  (Art.  22)  and 

[vector  (z'2  -  vx)\l(t%  -  tx) 
is  the  value  of  the  average  acceleration. 


214  THEORETICAL    MECHANICS. 

(Fig.  114),  and  the  velocity  after  the  short  interval  A/  by  0'B\  then 
(vector  A'B')  I  At 

is  an  approximate  value  of  the  required  acceleration.     The  approxi- 
mation is  closer  the  shorter  the  interval  At. 

250.  Exact  Value  of  Acceleration  at  an  Instant — The  true 
value  of  the  acceleration  at  the  instant  t  is  the  limit  approached  by 
the  above  approximate  value  as  the  interval  At  approaches  zero. 
That  is, 

acceleration  =  limit  [(vector  A'B') /At]. 

To  determine  this  limiting  value,  the  following  reasoning  may  be 

employed : 

Let  O'A'  and  O'B'  (Fig.  1 1 5)  represent  the  values  of  the  velocity 

at  the  beginning  and  end  of  At  respectively,  so  that  vector  A'B'  de- 
notes the  increment  of  velocity,  the 
curve  A'B'  being  the  hodograph  or 
curve  of  velocities  (Art.  245).  As  At 
approaches  o,  the  point  B'  approaches 
A'  along  the  curve  A'B'.  The  limit- 
ing direction  of 

(vector  A'B")  I  At 

is  that  of  the  tangent  to  the  curve  at 

Fig.  115.  A'.     Also,  the  limit  of  the  magnitude 

of  this  vector  quantity  is  the  derivative 

of  s'  with  respect  to  /,  if  s'  is  the  length  of  the  curve  A'B'  measured 

from  some  fixed  point  up  to  the  point  which  is  describing  it ;  that  is 

limit  [(length  A'B')/At]  =  ds'/dt. 

Consider  the  curve  A'B'  (Fig.  115)  to  be  described  by  a  moving 
particle  whose  position  at  every  instant  corresponds  to  that  of  the 
given  particle  describing  the  path  AB  ;  so  that  if  P  represents  the 
position  of  the  latter  at  any  instant,  and  the  vector  O'P'  its  velocity, 
P'  is  the  position  of  the  point  describing  the  hodograph  or  curve 
A'B'.     Evidently,  ds'/dt  is  the  velocity  of  the  point  P'.     Hence 

The  acceleration  of  a  point  describing  any  patJi  in  any  manner 
is  at  every  instant  equal  {in  magnitude  and  direction)  to  the  velocity 
of  the  point  describing  the  hodograph. 

The  direction  of  the  acceleration  is  always  inclined  toward  the 
concave  side  of  the  path,  but  nothing  further  can  in  general  be  said 


MOTION    IN   A    CURVED    PATH.  215 

as  to  its  direction  relative  to  the  path.  Neither  can  any  simple 
expression  be  given  for  its  magnitude  in  terms  of  the  coordinates  of 
the  path.* 

251.  Sudden  Change  of  Velocity. —  In  what  precedes  it  has 
been  assumed  that  when  the  velocity  changes  from  one  value  to 
another,  it  passes  through  a  continuous  series  of  intermediate  values, 
and  that  for  a  finite  change  of  velocity  a  finite  interval  of  time  is  re- 
quired. It  is  conceivable  that  this  condition  might  be  violated,  and 
that  a  particle  might  receive  an  instantaneous  increment  of  velocity. 
In  case  of  the  motions  actually  occurring  in  nature  there  is  no  reason 
to  believe  that  instantaneous  changes  of  velocity  occur.  Sudden 
changes  of  velocity  do,  however,  occur  and  will  be  treated  in  another 
place.      (See  Chapter  XVI.) 

Examples. 

1.  A  particle  describes  a  circle  of  radius  2  ft.  with  a  uniform  speed 
such  that  the  entire  circumference  is  described  in  1  sec.  Required 
(a)  the  velocity-increment  for  an  interval  of  o.  1  sec. ;  (b)  the  average 
acceleration  for  the  same  interval  (give  its  magnitude  and  the  angle 
its  direction  makes  with  that  of  the  initial  velocity). 

Ans.  {a)  7.766  ft. -per-sec. ;  (&)  77.66  ft. -per-sec.-per-sec. ;  angle 
=  1080. 

2.  A  particle  describes  a  circle  of  radius  r  with  uniform  speed  v. 
Required  {a)  the  magnitude  and  direction  of  the  increment  received 
by  the  velocity  while  the  particle  describes  one-fourth  of  the  circum- 
ference ;  (Jb)  the  magnitude  and  direction  of  the  average  acceleration 
for  the  same  interval. 

Ans.  (b)  (21/2/V)  (z/2/r),  inclined  1350  to  initial  velocity. 

3.  With  data  as  in  Ex.  2,  determine  the  average  acceleration  in 
any  finite  interval  At.  (Give  its  magnitude  and  the  direction  it 
makes  with  the  initial  velocity.) 

Let  AB  (Fig.  116)  represent  the  arc  described  in  time  At.  The 
velocity  at  the  beginning  of  the  interval  is  represented  by  a  vector 
O' A\  of  length  v,  perpendicular  to  the  radius  OA  ;  the  velocity  at 
the  end  of  the  interval  is  represented  by  a  vector  0'B\  also  of  length 

*  If  z'  is  regarded  as  a  vector  symbol  representing  the  velocity  in  direction 
as  well  as  in  magnitude,  and  if  A v  is  the  increment  of  the  vector  v  in  the  time 
At,  the  value  of  the  acceleration  may  be  written 

p  =  limit  [(vector  Av/At)~\  =  vector  (dv/dt). 
But  the  meaning  of  dv/dt  is  quite  different  from  that  of  the  same  expression 
when  v  denotes  the  speed. 


2l6  THEORETICAL    MECHANICS. 

v,  perpendicular  to  the  radius  OB.  The  velocity-increment  is  there- 
fore vector  A 'B',  and  the  average  accel- 
eration is 

(vector  A'B ')/At. 

The  length  A'B'  is  2V  sin  (A'O'B'/i). 
But  (expressing  angles  in  radians), 
angle  A'O'B'  =  angle  A  OB 
=  (arc  AB)/OA  =  (vAt)/r; 
hence 

length  A'B'  =  2V  sin  (v&tfzr), 
and    the   average   acceleration    has    the 
magnitude 

FlG"  II6"  ^  '£'/Af  =  (2Z//A0  sin  (vAt/zr). 

Its  direction  is  perpendicular  to  the  chord  AB. 

4.  Taking  the  result  reached  in  Ex.  3,  let  the  interval  At  approach 
the  limit  zero ;  determine  the  magnitude  and  direction  of  the  limit- 
ing value  of  the  average  acceleration.  Determine  thus  the  exact 
value  of  the  acceleration  at  any  instant. 

As  At  approaches  zero,  the  chord  AB  approaches  the  tangent  to 
the  circle  at  A,  and  the  direction  of  the  vector  A  ' B' ,  which  is  per- 
pendicular to  the  chord  AB,  approaches  AO.  Also,  since  the  angle 
A  OB  (  —  A'O'B')  approaches  zero  as  At  approaches  zero,  the  value 
of  the  angle  in  radians  may  in  the  limit  replace  the  sine ;  that  is, 
vAtfer  may  replace  sin  (y  AtJ2r).     Therefore 

limit  [A'B' /At]  =  limit  [(2v/At)(vAt/2r)]  ==  v2/r. 
That  is,  the  actual  acceleration  at  the  instant  when  the  particle  is  at 
A  is  directed  toward  the  center  of  the  circle  and  has  the  magnitude 
v2/r. 

5.  What  is  the  value  of  the  acceleration  of  a  particle  moving  as 
described  in  Ex.  1  ? 

Ans.  78.9    ft.-per-sec.-per-sec.  toward  the  center. 

6.  Draw  the  path  and  the  hodograph  for  the  motion  described 
in  Ex.  2.  Choosing  any  point  P  on  the  path,  find  the  corresponding 
point  P'  on  the  hodograph.  If  P  moves  with  the  moving  particle, 
what  is  the  velocity  of  P'  at  any  instant  ?  From  the  principle  stated 
in  Art.  250,  that  the  velocity  of  P'  is  equal  in  magnitude  and  direc- 
tion to  the  acceleration  of  P,  verify  the  result  of  Ex.  4. 

7.  A  particle  describes  the  perimeter  of  a  regular  hexagon  with 
uniform  speed  v.  It  receives  what  sudden  increment  of  velocity  as 
it  passes  an  angle  of  the  polygon  ? 

8.  What  sudden  increments  of  velocity  are  received  by  a  particle 
describing  the  perimeter  of  a  regular  polygon  of ;/  sides  with  uniform 
speed  v  ? 


MOTION   IN   A    CURVED    PATH.  217 


§  3.  Motion  and  Force. 

252.  Effect  of  a  Constant  Force. — The  effect  of  a  constant  force 
upon  the  motion  of  a  body  which  initially  is  either  at  rest  or  moving 
along  the  line  of  action  of  the  force  has  been  considered  in  Arts.  214 
and  215.  The  principle  there  stated  is  that  the  force  produces  a 
velocity-increment  proportional  directly  to  the  force  and  to  the  time 
during  which  it  acts,  and  inversely  to  the  mass  of  the  body.  The 
case  in  which  the  body  has  initially  a  velocity  not  parallel  to  the  force 
must  now  be  considered. 

If  a  moving  particle  be  acted  upon  by  a  force  whose  direction  is 
inclined  to  that  of  the  motion,  there  results  a  continuous  change  in 
the  direction  of  the  motion.  Thus,  suppose  a  particle  to  move  along 
the  straight  line  AB  (Fig.  117)  with  uniform  velocity,  not  being  acted 
upon  by  any  force  until  arriving  at  the 
point  B.  Let  it  then  be  acted  upon  by 
a  force  represented  by  the  vector  MN, 
which  remains  constant  in  magnitude 
and  direction  for  a  period  A/.  The 
particle  is  deflected  from  the  original 
straight  path  AB,  and  describes  some 
curve  BC.  The  velocity  thus  changes 
in  direction,  and  in  general  it  changes 

also  in  magnitude.  Representing  the  velocity  at  the  beginning  of 
the  interval  A/  by  a  vector  O'B'  (Fig.  117)  parallel  to  AB,  and 
the  velocity  at  the  end  of  the  interval  by  a  vector  O'C  parallel  to 
the  tangent  to  the  path  BC  at  C,  the  total  change  of  velocity  (or 
velocity-increment)  is  represented  by  the  vector  B'C.  The  magni- 
tude and  direction  of  this  velocity-increment  depend  upon  (1)  the 
magnitude  and  direction  of  the  force,  (2)  the  duration  of  the  interval 
A/,  and  (3)  the  mass  of  the  particle. 

(1)  The  direction  of  the  velocity-increment  agrees  with  that  of 
the  force,  and  its  magnitude  is  proportional  directly  to  the  magnitude 
of  the  force. 

(2)  Its  magnitude  is  proportional  directly  to  the  duration  of  the 
interval. 

(3)  Its  magnitude  is  proportional  inversely  to  the  mass  of  the 
particle. 


218 


THEORETICAL    MECHANICS. 


These  principles  may  be  stated  in  the  following  general  proposi- 
tion : 

A  constant  force  acting  upon  a  particle  gives  it  a  veto  city -incre- 
ment whose  direction  is  that  of  the  force,  and  whose  magnitude  is 
proportional  directly  to  the  force  and  to  the  time  during  which  it 
acts  and  inversely  to  the  mass  of  the  particle. 

This  proposition  is  identical  in  form  with  that  given  in  Art.  215. 
But  since  the  discussion  was  there  restricted  to  the  case  of  rectilinear 
motion,  the  velocity-increment  was  necessarily  parallel  to  the  line  of 
motion.  This  restriction  being  removed,  the  proposition  is  still  true, 
velocity  and  velocity-increment  being  regarded  as  vector  quantities. 
The  special  case  of  rectilinear  motion  is  included  in  the  general  state- 
ment. 

It  is  thus  seen  that  the  effect  of  a  constant  force  upon  the  motion 
of  a  particle  is  estimated  in  the  same  way,  whatever  the  initial  ve- 
locity of  the  particle.     The  velocity-increment  is  in  every  case  equal 


to  the  velocity  which  would  be  produced  by  the  same  force  acting 
for  the  same  time  on  the  same  particle  initially  at  rest.  The  final 
velocity  is  the  vector  sum  of  this  velocity-increment  and  the  initial 
velocity. 

The  path  of  the  particle  will  depend  both  upon  the  initial  velocity 
and  upon  the  magnitude  and  direction  of  the  force.  To  illustrate 
this,  compare  two  cases  in  which  the  force  has  the  same  value  while 
the  initial  velocity  has  different  values.  The  two  cases  are  shown  in 
Fig.  118.  In  each  case  the  initial  velocity  vx  is  represented  by  the 
vector  0'A'y  the  direction  being  the  same  in  both  cases  but  the 


MOTION    IN    A    CURVED    PATH.  2ig 

magnitude  being  much  greater  in  the  second  case  than  in  the  first. 
The  force  (represented  by  the  vector  MN)  has  the  same  magnitude 
and  direction  in  both  cases.  The  mass  of  the  particle  is  also  the 
same,  so  that  the  values  of  the  velocity-increment  in  a  given  time  are 
equal  in  the  two  cases.  The  velocity-increment  being  represented 
by  the  vector  A'B' ,  the  final  velocity  v.,  is  given  by  the  vector  O'B'. 
The  values  of  v%  in  the  two  cases  differ  both  in  magnitude  and  in 
direction.  The  two  paths  are  also  quite  different.  If  A  and  B  are 
the  initial  and  final  positions  of  the  particle,  the  tangent  at  A  is  par- 
allel to  v1  and  the  tangent  at  B  is  parallel  to  vv  The  curvature  of 
the  path  is  evidently  greater  in  the  first  case  than  in  the  second ; 
while  the  length  of  the  path  is  greater  in  the  second  case,  because 
the  speed  is  greater  throughout  the  interval. 

Examples. 

i.  A  body  is  projected  horizontally  with  a  velocity  of  50  ft. -per, 
sec. ,  after  which  it  is  acted  upon  by  the  constant  force  of  gravity. 
Determine  the  magnitude  and  direction  of  its  velocity  at  the  end  of 
o.  5  sec. ,  1  sec.  and  2  sec. 

[The  velocity-increment  due  to  gravity  is  to  be  computed  as  if 
the  body  fell  vertically  from  rest.     See  Art.  227.] 

2.  A  body  is  projected  with  a  velocity  of  50  ft.-per-sec.  in  a  direc- 
tion inclined  400  upward  from  the  horizontal.  Determine  the  mag- 
nitude and  direction  of  the  velocity  at  the  end  of  o.  5  sec. ,  1  sec.  and 
2  sec. 

Ans.  At  the  end  of  2  sec.  the  velocity  is  50  ft.-per-sec,  directed 
400  downward  from  the  horizontal  (very  nearly). 

3.  A  body  falling  vertically  at  the  rate  of  30  ft.  -per-sec.  receives 
a  blow  in  a  horizontal  direction  which,  if  the  body  had  been  at  rest, 
would  have  given  it  a  velocity  of  100  ft.-per-sec.  Determine  the 
magnitude  and  direction  of  the  velocity  immediately  after  the  blow. 

4.  With  data  of  Ex.  3,  determine  the  magnitude  and  direction  of 
the  velocity  1  sec.  after  the  blow,  assuming  gravity  to  be  the  only 
force  acting  on  the  body. 

An s.  1 17.8  ft.-per-sec.  inclined  310  56'  downward  from  the  hori- 
zontal. 

253.  Equation  of  Motion  for  Particle  Acted  Upon  by  a  Con- 
stant Force. — The  general  principle  above  stated  (Art.  252)  may  be 
expressed  by  an  equation  similar  to  that  given  in  Art.  216  for  the 
case  of  rectilinear  motion.  Let  the  increment  of  velocity  (vector 
A'B',  Fig.  118)  be  represented  by  Av}  and  the  force  (vector  MN, 
Fig.  118)  by  P.    Then  if  m  is  the  mass  of  the  particle  and  A/  the 


220  THEORETICAL    MECHANICS. 

time  in  which  the  constant  force  P  produces  the  velocity-increment 
Av,  the  general  principle  is  equivalent  to  the  equation 

Av  =  k(PAt/m). 

But  this  must  be  interpreted  as  a  vector  equation,  expressing  identity 
of  direction  as  well  as  equality  of  magnitude. 

Taking  units  of  force,  mass,  length  and  time  as  in  Art.  217  or 
Art.  218,  so  that  k  =  1,  the  equation  may  be  written 

P  =  m{AvjAt). 

But  Av/Aty.  the  increment  of  velocity  per  unit  time,  is  the  accel- 
eration (Art.  247).  Hence,  if  acceleration  is  designated  by/,  the 
equation  may  be  written 

P  =  mp,     or    p  =  P/m. 

Unless  otherwise  stated,  it  will  hereafter  be  assumed  that  units  are 
so  chosen  that  k  =  I. 

254.  Effect  of  a  Variable  Force.— A  particle  may  be  acted  upon 
by  a  force  which  varies  in  magnitude,  in  direction,  or  in  both  mag- 
nitude and  direction.  Consider  the  effect  of  such  a  variable  force 
upon  the  motion  of  the  particle. 

Let  the  vector  O'A'  (Fig.  114)  represent  the  velocity  of  the  par- 
ticle at  an  instant  /,  and  the  vector  O' Br  the  velocity  after  an  interval 
At.  Then  A ' B'  is  the  velocity-increment  received  by  the  particle 
during  the  interval. 

The  value  of  a  constant  force  which  would  produce  the  same 
velocity-increment  in  the  same  time  is  (by  Art.  253) 

m  (vector  A'B')/At. 

If  At  is  very  small,  the  actual  force  differs  little  from  this  value  during 
the  interval ;  and  the  true  value  of  the  force  at  the  beginning  of  the 
interval  is  the  limit  approached  by  this  approximate  value  as  At 
approaches  zero.  That  is,  if  P  is  the  value  of  the  force  and  /  that  of 
the  acceleration  at  the  beginning  of  the  interval, 

P  -=  limit  [in  (vector  A  'B')[At]  =  mp. 

Here  P  and  p  must  be  understood  as  vector  symbols,  and  the  equa- 
tion expresses  equality  both  in  magnitude  and  in  direction. 

Thus  the  equation  of  motion  has  the  same  form  when  the  force  is 
variable  as  when  it  is  constant. 


MOTION    IN   A    CURVED    PATH.  221 

255.  Resultant  of  Two  or  More  Forces  Acting  Simultaneously 
Upon  the  Same  Particle.  — Any  number  of  forces  acting  simultaneously 
upon  a  particle  are  equivalent,  in  their  effect  upon  the  motion,  to  a 
single  force.  For  the  acceleration  has  at  every  instant  a  definite 
value  p,  and  this  same  acceleration  would  result  from  the  action  of 
a  single  force  of  magnitude  mp,  agreeing  in  direction  with/. 

A  single  force  which,  acting  alone,  has  the  same  effect  as  several 
forces  acting  simultaneously,  is  called  their  resultant  (Art.  55). 

The  resultant  of  any  number  of  forces  acting  at  the  same  time 
upon  a  particle  is  a  force  equal  to  their  vector  sum. 

This  proposition,  for  the  case  of  two  forces,  expresses  the  principle 
of  the  parallelogram  of  forces.  Assuming  its  truth  in  case  of  two 
forces,  its  truth  for  any  number  of  forces  follows  immediately.  Its 
meaning  may  be  explained  as  follows : 

hetpl  be  the  acceleration  due  to  a  force  Px  acting  alone  upon  a 
given  particle,  and  pt  the  acceleration  due  to  a  force  P2  acting  alone 
upon  the  same  particle.  Then  if  the  particle  be  acted  upon  by  Px  and 
P.,  at  the  same  time,  its  acceleration  is  the  vector  sum  of  px  and  pt . 
If  the  particle  be  acted  upon  at  the  same  time  by  a  third  force  P3 
which,  acting  alone,  would  produce  acceleration  p^ ,  the  actual  accel- 
eration is  the  vector  sum  o{pl,p2  and/3.  And  similarly  for  any 
number  of  forces. 

Thus,  the  acceleration  due  to  the  concurrent  action  of  several 
forces  may  be  computed  in  either  of  two  ways, —  (1)  by  determining 
the  acceleration  due  to  each  acting  alone  and  taking  the  vector  sum 
of  these  separate  accelerations,  or  (2)  by  determining  the  acceleration 
due  to  a  single  force  equal  to  the  vector  sum  of  the  several  forces. 

This  principle  must  be  regarded  as  a  fundamental  law  of  Dynam- 
ics, derived  from  experience  and  not  deducible  from  any  simpler  law 
or  laws. 

256.  Equation  of   Motion  for   Particle  Acted  Upon   by  Any 

Number  of  Forces. —  The  equation  of  motion  may  be  written  in  the 

same  form  when  several  forces  act  upon  the  particle  as  when  only  a 

single  force  acts.     It  is  n 

&  P  =  mp, 

if  m  is  the  mass  of  the  particle,  p  its  acceleration  at  any  instant,  and 
/'the  vector  sum  of  all  forces  acting  upon  the  particle  at  that  instant. 

257.  Remarks  on  General  Equation  of  Motion. —  The  equation 

P=mp  .         .         .        .         .     (1) 


222  THEORETICAL    MECHANICS. 

may  thus  be  called  the  general  equation  of  motion  for  a  particle  acted 
upon  by  any  forces.  In  interpreting  it,  P  may  be  taken  to  mean  any 
single  force  and/  the  acceleration  due  to  that  force  ;  or  P  may  mean 
the  resultant  of  several  forces  and  p  the  acceleration  due  to  their 
combined  action.  In  either  case  it  is  to  be  understood  that  P  and/ 
are  vector  quantities  having  the  same  direction. 

It  must  also  be  remembered  that  unless  the  units  of  force,  mass, 
length  and  time  are  related  in  a  certain  manner  (Art.  217),  the  equa- 
tionwillbe  p  =  Kph)t     or     P==k'mp. 

The  constant  k  (or  k')  will  always  be  unity  provided  the  unit  force 
is  taken  as  that  force  which  will  give  to  the  unit  mass  the  unit  accel- 
eration. 

Thus,  equation  (1)  may  be  used  if  the  units  are  the  foot,  pound- 
mass,  second  and  poundal ;  or  the  centimeter,  gram,  second  and 
dyne.     (Art  217.) 

Or,  any  three  of  the  units  involved  may  be  chosen  arbitrarily, 
and  the  other  determined  in  such  a  way  as  to  make  the  constant  k 
unity.  In  particular,  the  "engineers'  kinetic  system"  of  units,  de- 
scribed in  Art.  218,  may  be  employed.  That  is,  in  using  the  equa- 
tion P  -—  mp,  we  may  take  the  pound-force  as  the  unit  force,  pro- 
vided the  mass  be  expressed  in  units,  one  of  which  is  equal  to  g 
pounds-mass.  If  the  mass  of  the  body  is  known  in  pounds,  its  value 
in  these  new  units  is  found  by  dividing  by  g.  Thus,  if  M  denotes 
the  mass  in  pounds,  the  equation  of  motion  will  be  P  =  (M/g)p. 

258.  Absolute  Unit  of  Force. — The  unit  force  which  was  usually 
employed  in  the  earlier  chapters  of  this  book  was  the  weight  of  a 
unit  mass.  The  most  precise  method  of  determining  the  magnitude 
of  a  force  practically  consists  in  comparing  it  with  this  unit.  Since, 
however,  the  weight  of  a  given  mass  changes  with  its  position  on  the 
earth's  surface,  forces  determined  experimentally  in  different  localities 
in  terms  of  this  unit  cannot  be  accurately  compared  unless  the  ratio 
between  the  weights  of  the  same  mass  in  the  two  localities  is  known. 

In  Art.  217  another  unit  of  force  was  defined,  —  the  force  which, 
acting  upon  the  unit  mass,  produces  the  unit  acceleration.  This  has 
been  called  an  absolute  unit,  since  it  is  not  dependent  upon  any  par- 
ticular force  such  as  the  weight  of  a  body,  nor  upon  the  place  at 
which  its  value  is  determined. 

A  force  whose  value  is  known  in  terms  of  the  weight  of  a  unit 


MOTION    IN    A    CURVED    PATH.  223 

mass  at  a  given  place  may  be  expressed  in  terms  of  the  absolute  unit 
if  the  value  of  g  (the  acceleration  due  to  gravity)  at  that  place  is 
known.  Thus,  a  force  equal  to  the  weight  of  m  pounds-mass  at  a 
certain  place  is  equal  to  mg  poundals,  if  g  is  the  acceleration  due  to 
gravity  at  that  place,  expressed  in  feet-per-second-per-second.  (Art. 
227.)  The  most  accurate  method  of  determining  the  value  of^at 
any  place  is  to  determine  the  time  of  vibration  of  a  pendulum  of 
known  length.      (Arts.  309  and  425.) 

This  definition  of  the  absolute  unit  of  force  depends  upon  the  ef- 
fect of  force  in  accelerating  mass.  Any  accurate  definition  of  force 
as  a  measurable  quantity  must  rest  upon  the  same  basis.  For  this 
reason  the  general  proposition  or  ' '  law  of  motion  ' '  stated  in  Art. 
252  is  often  regarded  as  a  definition  of  force  rather  than  as  a  proposi- 
tion about  forces.      (See  Art.  260.) 

259.  Newton's  Laws  of  Motion. — The  fundamental  laws  of  mo- 
tion, as  formulated  by  Newton,*  are  as  follows  : 

Law  I.  Every  body  continues  in  its  state  of  rest  or  of  uniform 
motion  in  a  straight  line,  except  in  so  far  as  it  may  be  compelled  by 
force  to  change  that  state. 

Law  II.  Change  of  motion  is  proportional  to  impressed  force, 
and  takes  place  in  the  direction  of  the  straight  line  in  which  the  force 
acts. 

Law  III.  To  every  action  there  is  always  an  equal  and  contrary 
reaction  ;  or,  the  mutual  actions  of  any  two  bodies  are  always  equal 
and  oppositely  directed. 

These  laws  include  all  the  principles  upon  which  the  foregoing 
discussions  have  been  based. 

Law  I  expresses  the  principle  stated  in  Arts.  31  and  213.  This 
is  often  called  the  law  of  inertia ;  inertia  being  defined  as  that  prop- 
erty of  matter  by  virtue  of  which  a  body  cannot  of  itself  change  its 
own  state  of  motion. 

Law  II  includes  in  its  meaning  the  principles  stated  in  Arts.  252- 
256.  In  order  that  this  may  be  clearly  understood,  some  explana- 
tion of  the  terms  employed  by  Newton  is  needed. 

The  word  motion  is  to  be  understood  as  meaning  what  is  defined 
as  momentum  (Art.  312);  that  is,  a  quantity  proportional  directly 
to  the  mass  of  a  body  and  to  its  velocity  at  any  instant.     Change  of 

*  These  laws  are  stated  nearly  in  the  language  of  Kelvin  and  Tait's 
"Natural  Philosophy,"  §g  244,  251,  261. 


224  THEORETICAL    MECHANICS. 

motion  is  therefore  change  of  momentum,  and  is  a  vector  quantity 
proportional  to  the  mass  and  to  the  change  of  velocity  or  velocity- 
increment.  Again,  impressed  force  must  be  understood  to  mean,  not 
simply  force  as  the  term  is  now  understood  and  has  been  used  in  the 
foregoing  discussions,  but  a  quantity  proportional  to  the  force  and  to 
the  time  during  which  it  acts.  Thus,  if  a  force  of  magnitude  P  acts 
for  an  interval  of  time  At  upon  a  body  of  mass  m,  and  if  Av  repre- 
sents the  resulting  velocity-increment,  the  law  asserts  that  m  Av  is 
proportional  to  P  At.  This  is  evidently  the  same  result  as  that  stated 
in  Art.  252  for  the  case  of  a  constant  force,  and  extended  to  the  case 
of  a  variable  force  in  Art.  254. 

That  Law  II  includes  also  the  law  of  composition  of  forces  acting 
simultaneously  (Art.  255)  is  not  immediately  evident  from  the  lan- 
guage in  which  it  is  expressed.  By  Newton,  however,  the  law  was 
doubtless  intended  to  imply  that  every  force  acting  upon  a  body 
produces  its  effect  independently  of  the  action  of  other  forces  upon 
the  same  body,  and  that  these  effects  (accelerations)  combine  accord- 
ing to  the  law  of  vector  addition.  * 

There  is  some  ground  for  the  opinion  that  the  law  of  composition 
of  forces  constitutes  an  independent  principle  among  the  fundamental 
laws  of  Dynamics.  Whether  this  be  the  correct  view  or  not,  it  is 
desirable  that  this  law  should  receive  explicit  statement.  Such  a 
statement  is  given  in  Art.  255. 

Law  III  has  been  stated  and  explained  in  Art.  35.  In  spite  of  its 
importance  and  its  simplicity,  its  true  meaning  has  often  been  missed. 

In  the  explanation  of  this  law  given  in  Art.  35,  the  word  action 
was  interpreted  as  meaning  force,  thus  limiting  the  law  to  a  state- 
ment regarding  the  forces  exerted  by  two  bodies  upon  each  other. 
By  Newton  the  law  was  applied  in  a  wider  sense,  as  indicated  by  the 
scholium  f  which  he  added  to  it.  This  scholium  is  here  omitted, 
since  its  full  meaning  cannot  be  explained  without  anticipating  the 
results  of  a  later  discussion.  The  law  will  for  the  present  be  under- 
stood as  referring  only  to  forces.  So  long  as  the  attention  is  confined 
to  the  motion  of  a  single  particle,  no  use  is  made  of  the  third  law. 
But  it  is  of  fundamental  importance  in  the  analysis  of  the  motions  of 
two  or  more  particles  which  exert  forces  upon  one  another.  The 
importance  of  the  law  in  Statics  has  already  been  seen  in  Chapter  VI. 

*See  Kelvin  and  Tait's  "Natural  Philosophy,"  §|  254,  257. 
t Thomson  and  Tait's  ''Natural  Philosophy,"  \  263. 


MOTION    IN    A    CURVED    PATH.  225 

Examples. 

1.  A  constant  force  acting  upon  a  body  of  mass  40  lbs.  for  3  sec. 
changes  its  velocity  from  20  ft.-per-sec.  north  to  80  ft.-per-sec.  east. 
Required  the  magnitude  of  the  force  in  poundals. 

Ans.    1099  poundals. 

2.  If  the  same  force  continues  to  act,  what  will  be  the  velocity  of 
the  body  after  3  more  sec.  ? 

Ans.    161. 3  ft.-per-sec.     S.  820  52'  E. 

3.  A  particle  of  5  lbs.  mass  is  acted  upon  by  two  constant  forces 
at  right  angles  to  each  other;  one  equal  to  the  weight  of  1  lb.,  the 
other  equal  to  20  poundals.  What  single  force  would  produce  the 
same  effect?  Ans.  Magnitude  of  force  =  37.9  poundals. 

4.  In  Ex.  3,  if  the  particle  has  initially  a  velocity  of  20  ft.-per- 
sec.  in  a  direction  opposite  to  that  of  the  lesser  force,  determine  its 
velocity  after  2  sec. 

Ans.    17.6  ft.-per-sec.  at  angle  of  470  2  with  initial  velocity. 

5.  A  particle  whose  mass  is  250  gr.  is  acted  upon  by  two  con- 
stant forces  whose  directions  are  inclined  to  each  other  at  an  angle  of 
6o°.  One  force  is  equal  to  the  weight  of  the  particle,  the  other  is 
equal  to  50,000  dynes.  If  the  velocity  at  a  certain  instant  is  600 
c.m.-per-sec.  in  a  direction  bisecting  the  angle  between  the  forces, 
determine  the  velocity  3  sec.  later. 

6.  In  Ex.  5,  what  single  force  may  replace  the  given  forces? 

7.  If  a  body  of  38  lbs.  mass  describes  a  circle  of  40  ft.  diameter 
at  the  uniform  rate  of  30  ft.-per-sec,  what  is  the  magnitude  and 
direction  of  the  resultant  force  acting  upon  it  at  any  instant  ? 

[Use  result  of  Ex.  4,  following  Art.  251.] 

8.  A  body  of  mass  m  describes  a  circle  of  radius  r  at  the  uniform 
rate  of  v  ft.  -per-sec.  Required  the  magnitude  and  direction  of  the 
resultant  force  acting  upon  it  at  any  instant. 

260.  Laws  of  Motion  Regarded  as  Definitions  of  Force  and 
Mass. —  In  the  foregoing  exposition  of  the  laws  of  motion,  it  has 
been  tacitly  assumed  that  force  and  mass  are  quantities  each  of  which 
is  measurable  independently  of  the  other  ;  so  that  either  or  both  may 
be  made  fundamental  in  the  system  of  units  which  we  may  choose 
to  adopt.  (Art.  216.)  But  when  it  is  attempted  to  define  these 
quantities  with  scientific  accuracy,  it  appears  to  be  impossible  to 
avoid  making  use  of  the  laws  of  motion  themselves.  It  has  already 
been  shown  that  the  second  law  furnishes  a  means  of  denning  the  unit 
force  in  an  exact  manner  when  the  unit  mass  has  been  chosen  (Art. 
258).  A  closer  analysis  shows  that  exact  definitions  of  both  force 
and  mass  as  measurable  quantities  are  supplied  by  the  laws  of  motion. 

15 


226  THEORETICAL    MECHANICS. 

Notice  first  that  if  it  is  possible  to  subject  two  different  bodies  to 
the  action  of  equal  forces,  the  second  law  furnishes  a  definite  com- 
parison of  their  masses.  Let  bodies  whose  masses  are  mx  and  m2  be 
acted  upon  by  equal  forces  of  magnitude  P,  and  let  px ,  p2  be  their 
accelerations.     Then  (Art.  253) 

mxpx  =  m2p2  ; 

and  if  px  and  p2  can  be  measured,  the  ratio  of  the  masses  is  at  once 
known.  If  a  third  mass  m^  be  acted  upon  by  a  force  equal  to  P, 
the  ratio  of  its  mass  to  that  of  mx  or  mt  becomes  known  if  its  ac- 
celeration p3  is  measured.      In  fact, 

«i  =  (A  /A  )«i ;   ^3  =  (a  /a)**i  ; 

and  if  mx  is  chosen  as  the  unit  mass,  the  numerical  values  of  m2  and 
m3  are  known.  Thus  the  mass  of  any  body  whatever  can  be 
expressed  in  terms  of  mx  as  a  unit  if  it  can  be  determined  what  ac- 
celeration it  will  have  when  acted  upon  by  a  force  equal  to  P. 

Evidently,  however,  it  is  not  necessary  always  to  apply  a  force  of 
the  same  magnitude  P.  The  masses  of  two  particles  may  be  com- 
pared by  comparing  the  accelerations  produced  in  them  by  equal 
forces  of  any  magnitude  whatever  ;  and  two  masses  may  be  compared 
indirectly,  by  comparing  each  independently  with  a  third  mass,  forces 
of  different  magnitudes  being  used  in  the  two  comparisons. 

Now  our  conception  of  the  laws  of  motion  implies  that  all  such 
comparisons  of  two  masses  mx}  m2,  direct  and  indirect,  with  any 
values  whatever  of  the  forces  whose  effects  are  measured  in  making 
the  comparisons,  lead  to  the  same  value  of  the  ratio  mxjm2.  This 
must  be  true  if  the  science  based  upon  these  laws  is  self- consistent. 
Assuming  it  to  be  true,  we  have  a  method  of  defining  mass  as  a 
measurable  magnitude  in  a  consistent  and  exact  manner. 

But  is  it  possible  to  define  the  meaning  of  '  ■  equal  forces  applied 
to  different  bodies ' '  without  making  use  of  the  notion  of  mass  itself? 
The  answer  to  this  question  is  that  such  a  definition  is  supplied  by 
the  third  law,  —  the  equality  of  action  and  reaction.* 

*  Maxwell's  answer  is  that  the  desired  object  may  be  accomplished  by- 
conceiving  the  forces  applied  by  means  of  an  elastic  string,  one  end  of  which 
is  attached  to  each  of  two  or  more  bodies  in  turn.  The  other  end  is  pulled 
in  such  a  way  that  the  string  elongates  by  a  certain  amount  which  is  the  same 
in  every  case  and  is  kept  constant  during  each  experiment.     Assuming  the 


MOTION    IN   A    CURVED    PATH.  227 

The  acceleration  of  a  particle  at  any  instant  is  due  to  the  influ- 
ence of  other  particles.  The  "action"  of  a  particle  A  upon  a  par- 
ticle B  which  causes  acceleration  of  B  is  called  the  ' '  force  exerted 
by  A  upon  B. ' '  The  third  law  of  motion  asserts  that  A  and  B  influ- 
ence each  other  mutually,- so  that  the  force  exerted  by  A  upon  B 
and  the  force  exerted  by  B  upon  A  are  at  every  instant  equal  in 
magnitude  and  opposite  in  direction. 

Let  the  masses  of  the  two  particles  be  mx  and  m2.  Let  that 
part  of  the  acceleration  of  A  which  is  due  to  B  be  px ,  and  let  that 
part  of  the  acceleration  of  B  which  is  due  to  A  be  p2.  The  "force" 
exerted  by  B  upon  A  is  then  measured  by  mxpx,  and  the  force  ex- 
erted by  A  upon  B  by  mtp2.  By  the  third  law,  these  are  equal  in 
magnitude ;  that  is, 

mxpx  =  mtp2,     or     mjmx  =  pjp2. 

This  equation  may  be  regarded  as  the  definition  of  the  ratio  of  tlie 
masses  of  the  two  bodies.     Or,  in  words, 

The  masses  of  any  two  bodies  are  in  the  inverse  ratio  of  the  ac- 
celerations they  give  each  other  at  any  instant. 

This  conception  of  the  meaning  of  mass  and  of  force,  and  of  the 
significance  of  the  laws  of  motion,  may  be  embodied  in  the  following 
statements  : 

(1)  The  acceleration  of  a  particle  is  always  due  to  the  influence 
of  other  particles. 

(2)  The  actual  acceleration  of  a  particle  is  at  every  instant  equal 
to  the  vector  sum  of  accelerations  due  to  the  individual  particles 
which  are  influencing  it. 

(3)  The  accelerations  which  two  particles  experience  by  reason  of 
their  mutual  influence  are  at  every  instant  oppositely  directed  along 
the  line  joining  them. 

(4)  The  masses  of  two  particles  are  in  the  inverse  ratio  of  the 
accelerations  they  give  each  other.     This  is  a  definition  of  mass. 

(5)  The  product  of  the  mass  of  a  particle  A  into  that  part  of  its 
acceleration  which  is  due  to  another  particle  B  is  the  measure  of  the 
force  exerted  by  B  upon  A.     This  is  a  definition  of  force. 

elastic  properties  of  the  string  to  remain  constant,  it  may  be  assumed  that  the 
magnitude  of  the  force  is  the  same  in  every  case.  ("Matter  and  Motion," 
Chapter  III.)  Even  if  no  actual  string  can  be  found  for  which  this  will  be 
true,  this  may  perhaps  still  be  regarded  as  a  satisfactory  method  of  explaining 
ideally  what  is  meant  by  equal  forces. 


228  THEORETICAL    MECHANICS. 

The  first  of  these  statements  is  virtually  Newton's  first  law  ;  for 
it  implies  that  if  a  particle  were  free  from  the  influence  of  other  par- 
ticles (that  is,  from  the  action  of  forces)  it  would  have  no  acceleration. 

The  second  statement  is  the  law  of  composition  of  forces,  usually 
regarded  as  included  in  Newton's  second  law.  The  second  law  as 
applied  to  the  effect  of  a  single  force  is  included  in  statements  (4) 
and  (5).      Newton's  third  law  is  included  in  (3),  (4)  and  (5).* 

261.  Particles  and  Bodies. — The  geometrical  analysis  of  motion 
in  the  present  Chapter  and  in  Chapter  XII  referred  to  a  particle. 
The  laws  of  motion,  although  stated  by  Newton  as  applying  to 
"bodies,"  have  here  been  applied  only  to  the  ideal  bodies  called 
particles.  The  propositions  at  the  end  of  Art.  260  were  also  stated 
as  applying  to  particles.  These  particles  have  been  regarded  as  pos- 
sessing finite  mass  while  occupying  no  finite  volume.  That  such  a 
particle  exists,  either  in  isolation  or  associated  with  other  particles  so 
as  to  constitute  one  of  the  ' '  bodies  ' '  of  our  experience,  we  are  not 
justified  in  asserting.  It  is,  then,  pertinent  to  inquire  what  applica- 
bility this  abstract  theory  has  to  the  actual  bodies  of  nature. 

To  answer  this  question  it  is  necessary  to  anticipate  the  results 
which  are  reached  when  the  theory  of  the  motion  of  a  particle  is 
extended  to  systems  of  connected  particles.  Assuming  the  accelera- 
tion of  every  particle  to  be  in  accordance  with  Newton's  second  law 
as  above  explained,  and  assuming  the  influence  of  the  particles  upon 
one  another  to  be  in  accordance  with  Newton's  third  law,  it  is 
possible  to  deduce  certain  general  laws  governing  the  motion  of  con- 
nected systems  of  particles.  These  laws  are  found  to  describe  with 
great  accuracy  the  actual  motions  of  natural  bodies. 

Moreover,  it  is  immaterial  whether  a  body  is  regarded  as  made 
up  of  a  finite  number  of  particles,  each  of  finite  mass,  or  whether  it  is 
regarded  as  occupying  space  continuously  (Art.  5).  If  by  particle 
we  understand  a  portion  of  matter  whose  volume  and  mass  are  both 
vanishingly  small,  the  foregoing  theory  of  the  motion  of  a  particle 
may  be  applied  if  it  be  assumed  that  the  force  acting  upon  any  par- 
ticle becomes  vanishingly  small  with  the  mass.     To  extend  the  theory 


*  For  a  critical  analysis  of  the  laws  of  motion  along  the  lines  of  the  above 
discussion,  the  following  works  may  be  consulted:  "  Science  of  Mechanics," 
by  Dr.  Ernst  Mach,  English  translation  by  Thomas  J.  McCormack.  "Gram- 
mar of  Science,"  by  Karl  Pearson.  "Theoretical  Mechanics,"  by  A.  E.  H. 
Love. 


MOTION    IN    A    CURVED    PATH.  229 

to  the  motion  of  a  body  of  finite  size  involves,  on  this  hypothesis,  a 
process  of  integration,  but  the  general  laws  deduced  do  not  differ 
from  those  resulting  from  the  other  assumption. 

Newton's  first  and  second  laws  are  not  intelligible  as  applied  to  a 
body  of  finite  size,  unless  it  be  assumed  that  all  parts  of  the  body 
have  at  every  instant  equal  and  parallel  velocities.  This  is,  indeed, 
a  possible  state  of  motion,  but  its  actual  occurrence  is  very  excep- 
tional. It  is  not  possible  to  describe  simply  the  way  in  which  a  body 
would  move  if  not  influenced  by  other  bodies  ;  still  less  can  a  simple 
general  statement  be  made  describing  the  effect  of  a  single  external 
force  in  accelerating  the  various  parts  of  a  body. 

One  of  the  results  reached  in  the  theory  of  the  motion  of  con- 
nected systems  of  particles  is  that  there  is  in  every  such  system 
a  certain  point  (the  center  of  mass)  whose  motion  is  accurately 
described  by  Newton's  first  and  second  laws,  if  the  whole  mass  be 
regarded  as  concentrated  at  that  point  and  acted  upon  by  forces 
equal  in  magnitude  and  direction  to  the  external  forces  which  are 
actually  applied  to  the  system. 

§  4.   Simultaneous  Motions. 

262.  Meaning  of  Simultaneous  Motions. —  The  actual  motion 
of  a  particle  is  sometimes  regarded  as  the  resultant  of  two  or  more 
component  motions  occurring  simulta- 
neously. 

Thus,  if  the  total  displacement  dur- 
ing a  given  interval  is  AB  (Fig.  119), 
the  particle  may  be  regarded  as  receiv- 
ing simultaneously  the  displacements 
AC>  CB,  even  though  the  actual  dis- 
placement follows  a  path  A  MB  not  con- 
taining the  point  C. 

Similarly,  if  the  actual  velocity  at  a 
given  instant  is  represented  by  the  vec- 
tor A'B'  (Fig.  120),  the  particle  may  be  regarded  as  having  simul- 
taneously two  velocities  represented  by  A'C,  CB'.  And  if  the 
actual  acceleration  at  any  instant  is  represented  by  the  vector  A" B" 
(Fig.  121),  the  particle  may  be  regarded  as  having  simultaneously 
two  accelerations  represented  by  A" C\  C" B" . 


23O  THEORETICAL    MECHANICS. 

Any  actual  displacement,  velocity  or  acceleration  may  thus  be 
regarded  as  the  resultant  of  any  number  of  "simultaneous"  com- 
ponents which  satisfy  the  condition  that  their  vector  sum  is  equal  to 
the  given  resultant. 

C" 


263.  Resolution  Into  Simultaneous  Motions  an  Arbitrary 
Process. —  Obviously  a  particle  cannot  at  the  same  time  describe  two 
different  paths  ;  neither  can  it  have,  at  the  same  instant,  two  different 
velocities  or  accelerations. 

The  motion  has  at  every  instant  a  definite  direction  and  a  def- 
inite speed,  and  the  actual  velocity  is  represented  by  a  definite 
vector.  This  may  be  expressed  as  the  sum  of  several  component 
vectors,  but  it  is  only  by  an  arbitrary  use  of  language  that  these  com- 
ponent vectors  can  be  said  to  represent  velocities  actually  possessed 
by  the  particle  at  the  same  instant. 

264.  Effects  of  Simultaneous  Forces  May  Be  Estimated  Sep- 
arately.—  One  case  in  which  it  may  be  advantageous  to  regard  the 
actual  motion  of  a  particle  as  the  resultant  of  two  or  more  simulta- 
neous motions  is  that  in  which  the  particle  is  acted  upon  by  two  or 
more  forces  such  that  the  effect  of  each  acting  alone  admits  of  simple 
determination.  From  the  law  of  composition  of  forces  (Art.  255) 
the  actual  acceleration  is  the  vector  sum  of  the  accelerations  which 
would  result  from  the  several  forces  acting  singly.  Hence  these 
component  accelerations  may  be  computed  separately  and  then  com- 
bined. This  method  may  be  carried  further,  and  applied  to  the 
computation  of  the  velocity-increment  and  the  displacement  due  to 
the  simultaneous  action  of  several  forces,  constant  or  variable,  during 
any  interval. 

265.  Velocity-Increment  Due  to  Several  Forces  Acting  Simul- 
taneously.—  Since  acceleration  is  by  definition  (Art.  247)  the  incre- 
ment of  velocity  per  unit  time,  and  since,  by  the  law  of  composition 
of  forces,  the  acceleration  at  each  instant  is  the  vector  sum  of  the 


MOTION    IN    A    CURVED    PATH.  23I 

accelerations  which  would  result  from  the  separate  action  of  the  forces, 
it  follows  immediately  that  the  actual  increment  of  velocity  prodiiced 
in  any  interval  is  the  vector  sum  of  tJie  velocity-increments  which 
would  result  from  the  separate  action  of  the  forces  during  the  same 
interval. 

If  at  the  beginning  of  a  certain  interval  the  velocity  is  v0,  and 
if  during  the  interval  the  particle  is  acted  upon  by  forces  Pi%  P%i 
.  .  .  ,  constant  or  variable,  which  separately  would  produce  ve- 
locity-increments v1 ,  yt ,  .  .  .  ,  the  velocity  at  the  end  of  the 
interval  is  the  vector  sum  of  v0 ,  v1 ,  v2 , 

266.  Displacement  Due  to  Several  Forces  Acting  Simultane- 
ously.—  Since  the  displacement  of  a  particle  during  any  interval 
depends  upon  the  magnitude  and  direction  of  its  velocity  at  every 
instant  throughout  that  interval,  the  total  displacement  may  be  com- 
puted as  the  vector  sum  of  the  displacements  due  to  any  components 
into  which  the  actual  velocity  may  be  resolved.  Thus,  suppose  a 
particle,  having  at  the  beginning  of  a  certain  interval  At  some  definite 
velocity  vQ ,  to  be  acted  upon  during  the  interval  by  a  force  P.  The 
position  of  the  particle  at  the  end  of  the  interval  may  be  determined 
as  follows : 

Determine  separately  (a)  the  displacement  in  the  time  At  due  to 
the  initial  velocity  if  no  force  acted,  and  (£)  the  displacement  in  the 
time  At  due  to  the  force  P  acting  on 
the  particle  initially  at  rest.  The  vector 
sum  of  these  displacements  is  the  actual 
resultant  displacement.  This  vector, 
drawn  from  the  initial  position  of  the 
particle,  determines  its  position  at  the 
end  of  the  interval. 

Thus,  the  displacement  due  to  the  Fig.  122. 

initial  velocity  v0  is  a  vector  whose  di- 
rection is  that  of  v0  and  amose  length  is  v0  At.     Represent  this  by 
AB  (Fig.  122).     If  BC  represents  the  displacement  which  the  par- 
ticle would  receive  in  the  time  At  if  initially  at  rest  and  acted  upon 
by  the  force  P,  the  vector  A  C  represents  the  resultant  displacement. 

The  path  followed  by  the  particle  from  A  to  C  is  some  curve 
AMC  tangent  to  AB  at  A. 

If  two  forces  Pl  and  P2  act  upon  the  particle  during  the  interval 
At,  the  total  displacement  may  be  computed  as  the  vector  sum  of 


232  THEORETICAL    MECHANICS. 

three  components :  a  component  v0  At  having  the  direction  of  the 
initial  velocity  v0,  a  component  equal  to  the  displacement  of  the  par- 
ticle if  initially  at  rest  and  acted  upon  by  the  force  Px  alone,  and  a 
component  equal  to  the  displacement  due  to  P2  acting  alone  on  the 
particle  initially  at  rest. 

The  same  method  may  be  applied,  whatever  the  number  of  the 
forces. 

Examples. 

1.  A  particle  is  projected  horizontally  with  a  velocity  of  20  ft.- 
per-sec,  after  which  it  is  acted  upon  by  no  force  except  gravity. 
Determine  its  velocity  and  its  position  after  o.  1  sec,  0.2  sec,  0.5 
sec,  2  sec. 

Ans.  Let  v  =  velocity,  <f>  =  angle  between  vsrnd  the  horizontal, 
r  =  distance  from  starting  point,  6  =  angle  between  r  and  the  hori- 
zontal. At  the  end  of  2  sec.  v  =  67.4  ft. -per-sec ,  (f>  =  72°  45', 
r=  75-8  ft.,  0=  58°  9'. 

2.  A  particle  is  projected  horizontally  with  a  velocity  v0,  after 
which  it  is  acted  upon  by  no  force  except  gravity.  Determine  its 
velocity  and  its  position  after  /  seconds. 

3.  A  particle  is  projected  with  a  velocity  of  100  ft. -per-sec.  in  a 
direction  inclined  20°  upward  from  the  horizontal.  Determine  its 
velocity  and  its  position  after  o.  1  sec ,  o.  5  sec ,  2  sec. 

267.  Relativity  of  Motion. —  The  motion  of  a  body  is  always 
estimated  with  reference  to  some  standard  body  which  is  assumed  to 
be  "fixed."  Thus,  in  all  ordinary  practical  problems,  the  earth  is 
regarded  as  a  fixed  body,  of  unchanging  dimensions  ;  the  displace- 
ment and  velocity  of  any  terrestrial  body  are  estimated  as  they  appear 
to  an  observer  on  the  earth.  If  a  body  moves  from  A  to  Bf  these 
being  two  points  fixed  upon  the  earth,  the  straight  line  AB  is  its 
total  displacement  when  the  earth  is  the  standard  body  with  refer- 
ence to  which  motions  are  specified.  If,  however,  the  rotation  of 
the  earth  upon  its  axis  is  considered,  the  displacement  has  a  very 
different  value.  Similarly,  the  velocity  and  acceleration  of  a  par- 
ticle at  any  instant  have  different  values  if  different  reference  bodies 
are  selected. 

In  the  foregoing  discussions,  and  in  those  that  follow  unless  the 
contrary  is  explicitly  stated,  it  is  to  be  understood  that  only  a  single 
reference  body  is  considered.  When  the  motions  of  different  bodies 
or  particles  are  discussed,  or  when  the  motion  of  a  particle  is  regarded 
as  made  up  of  components  which  follow  certain  rules  of  combination, 


MOTION    IN    A    CURVED    PATH.  233 

all  these  motions  are  estimated  with  reference  to  the  same  standard 
body. 

268.  Motion  of  a  Particle  Referred  to  Two  Different  Bodies.— 

In  some  cases  the  study  of  the  motion  of  a  particle  is  aided  by 
regarding  it  from  two  different  points  of  view, —  the  motion  being 
referred  first  to  one  standard  body  and  then  to  another.  This  is 
illustrated  by  the  following  simple  case  : 

Let  a  particle  P slide  along  a  tube  AB  (Fig.  123),  while  the  tube 
itself  is  in  motion.  Let  the  particle  move  from  A  to  B  while  the 
tube  moves  from  the  position  A'B'  to  the  position  A"B";  AB  being 
in  every  position  parallel  to  its  original 

position  A'B',  and  the  ends  of  the  tube  £»  2?" 

describing  the  straight  lines  A'A",B'B". 
The  total  displacement  of  the  particle 
is  from  A'  to  B  ".  The  actual  displace- 
ment may  follow  any  curve  or  broken 

line  lying  between  A '  B' B"  and  A  'A "  B" .    A'  **' 

.  Fig.  123. 

If  the  motion  of  the  particle  along  the 

tube  from  A   to  B,  and  the  motion  of 

the  tube  from  A'B'  to  A"B",  both  take  place  uniformly,  the  actual 

displacement  will  be  along  the  vector  A'B";   but  the  same  total 

displacement  of  the  particle  may  occur  in  many  other  ways. 

It  will  be  observed  that  in  thus  analyzing  the  motion  of  P  two 
different  reference  bodies  are  employed  in  estimating  the  motion. 
Thus,  AB  (or  A'B'  or  A"B")  is  the  displacement  of  the  particle  with 
respect  to  the  tube  ;  while  A' A"  (or  B'B")  is  the  displacement  of  the 
tube  with  respect  to  some  "fixed"  body  not  specified ;  and  A'B"  is 
the  displacement  of  the  particle  with  respect  to  this  same  "fixed" 
body.  Of  these  three  displacements  the  third  is  seen  to  be  equal  to 
the  vector  sum  of  the  first  and  second,  and  the  actual  displacement 
of  the  particle  with  respect  to  the  "  fixed  "  reference  body  is  therefore 
regarded  as  the  resultant  of  the  two  "simultaneous"  displacements 
AB',  B'B". 

Let  it  now  be  assumed  that  the  particle  P  moves  along  the  tube 
AB  at  a  uniform  rate,  while  the  tube  moves  in  the  direction  A' A" 
also  at  a  uniform  rate.  If  the  particle  is  at  the  point  A  of  the  tube 
when  the  tube  is  at  A'B',  and  if  it  moves  from  A  to  B  while  the 
tube  moves  from  A'B'  to  A" B" ,  the  actual  path  of  the  particle 
is   the  vector   A'B".      If  this   total   motion    occupies   At   seconds, 


234  THEORETICAL    MECHANICS. 

the  actual  velocity  of  the  particle  throughout  the  motion  has  the 

value  (vector  A'B")/At. 

This  is  obviously  the  vector  sum  of  two  components, 

(vector  AB)/At    and     (vector  B'B")/At, 

of  which  the  former  represents  the  velocity  of  the  particle  with  re- 
spect to  the  tube,  and  the  latter  the  velocity  of  the  tube  with  respect 
to  the  "fixed"  reference  body. 

Hence  the  "  actual"  velocity  of  the  particle  (t.  e.,  its  velocity  with 
respect  to  the  assumed  ' '  fixed ' '  reference  body)  may  be  regarded  as 
the  resultant  of  two  simultaneous  velocities,  one  of  which  is  its 
velocity  with  respect  to  the  tube  and  the  other  is  the  velocity  of  the 
tube. 

An  analysis  like  the  above  is  useful  in  many  cases  in  which  the 
displacement  and  velocity  of  a  particle  may  easily  be  estimated  with 
reference  to  a  body  which  is  itself  in  motion. 

Examples. 

i.  A  straight  tube  AB,  i  met.  long,  moves  in  such  a  way  that 
every  point  has  a  velocity,  relative  to  the  earth,  of  12  c.  m. -per-sec. 
in  a  direction  inclined  450  to  AB.  A  particle  slides  uniformly  from 
A  to  B  in  3  sec.  Required  (a)  the  velocity  of  the  particle  relative 
to  the  tube ;  (b)  its  velocity  relative  to  the  earth ;  (c)  its  total  dis- 
placement relative  to  the  earth  while  it  slides  from  A  to  B.  [Give 
direction  as  well  as  magnitude  of  each  of  the  required  vector  quan- 
tities. ] 

2.  A  stream  of  water  flows  uniformly  at  the  rate  of  2  miles  per 
hour.  A  boat  is  rowed  in  such  a  way  that  in  still  water  its  velocity 
would  be  5  ft.  -per-sec.  in  a  straight  line.  If  headed  directly  across 
the  current,  (a)  what  is  its  velocity  referred  to  the  earth?  (b)  If 
the  stream  is  3,000  ft.  wide  and  the  boat  starts  from  one  shore,  where 
will  it  strike  the  opposite  shore  ? 

3.  If,  in  Ex.  2,  the  boat  is  headed  up-stream  at  an  angle  of  300 
with  the  shore,  answer  questions  (a)  and  (b). 

Arts,  (a)  2.86  ft. -per-sec. ,  290  12'  up-stream,  (b)  0.317  miles 
up-stream. 

4.  In  Ex.  2,  how  must  the  boat  be  headed  in  order  to  strike  the 
opposite  shore  as  near  as  possible  to  the  starting  point  ? 

Ans.   350  55'  up-stream. 

5.  Show  that  the  current  does  not  affect  the  time  of  crossing  the 
stream. 


MOTION    IN    A    CURVED    PATH.  235 

6.  A  straight  tube  AB,  4  ft.  long,  rotates  about  the  end  A  at 
the  uniform  rate  of  1  revolution  per  sec.  A  particle  slides  along 
the  tube  from  B  toward  A  at  the  rate  of  10  ft.-per-sec.  Determine 
the  velocity  of  the  particle  relative  to  the  earth  when  it  is  midway 
between  A  and  B. 

Ans.  If  Cis  the  position  of  the  particle,  its  velocity  is  16.06  ft.- 
per-sec.  in  a  direction  making  the  angle  510  30'  with  CA. 


CHAPTER    XV. 


PLANE    MOTION    OF    A    PARTICLE. 

§  i .  Methods  of  Specifying  Motion  in  a  Plane. 

269.  Restriction  to  Plane  Motion. —  The  principles  developed 
in  Chapter  XIV  apply  to  any  motion  of  a  particle  in  space.  In  con- 
sidering the  methods  of  applying  these  principles  to  the  solution  of 
particular  problems,  the  discussion  will  be  limited  mainly  to  the  case 
in  which  the  path  of  the  particle  lies  in  a  plane.  It  is  needful  now 
to  consider  how  the  motion  can  conveniently  be  specified  in  this 
restricted  case. 

270.  Coordinates  of  Position. —  If  the  motion  of  a  particle  is 
restricted  to  a  plane,  its  position  at  any  instant  may  be  specified  by 

the  values  of  two  quantities.  Any  two 
quantities  serving  this  puroose  are  coor- 
dinates of  position. 

Of  the  various  possible  systems  of 
coordinates,  the  most  useful  are  the  rect- 
angular system  and  the  polar  system. 

Rectangular  coordinates. — Let  OX 
and  OY  (Fig.  124)  be  any  two  fixed 
lines  at  right  angles  to  each  other,  called 
axes  of  coordinates,  and  let  P  be  the 
position  of  the  moving  particle  at  any  instant.  Then  the  distances 
PN  and  PM,  measured  parallel  to  OX 
and  O  Y  respectively,  are  the  rectangu- 
lar coordinates  of  P,  and  will  be  denoted 
by  x  and  y. 

As  the  particle  moves,  x  and  y  vary. 
If  x  and  y  are  known  functions  of  the 

time,  the  position  of  P  is  known  at  every   Q  J[ 

instant.  Fig.  125. 

Polar  coordinates.  —  Let   the   fixed 
point  O  and  the  fixed  line  OX  (Fig.   125)   be  taken   as  pole  and 
initial  line  respectively  ;  then  the  angle  POX  (or  0)  and  the  length 
OP  (or  r)  are  the  polar  coordinates  of  the  particle  P.     As  P  moves, 


y 

N 

X 

.. .  TD 

1 

j- 

y 

0 

M 

[    X 

Fig.  124. 


PLANE    MOTION    OF    A    PARTICLE. 


237 


its  position  is  known  at  every  instant  if  r  and  6  are  known  functions 
of  the  time. 

271.  Velocity  and  Acceleration  Each  Given  by  Two  Compo- 
nents.—  Any  vector  quantity  lying  in  a  given  plane  is  known  when 
its  components  in  two  given  directions  are  specified.  In  analyzing 
the  motion  of  a  particle  in  a  plane,  the  displacement,  velocity  and 
acceleration  may  each  be  regarded  as  the  resultant  (or  vector  sum)  of 
two  components,  the  directions  of  resolution  being  chosen  at  pleasure. 
The  particular  directions  chosen  will  be  determined  by  convenience, 
depending  upon  the  nature  of  the  problem  under  consideration.  In 
nearly  every  case,  however,  the  two  directions  of  resolution  will  be 
taken  at  right  angles  to  each  other. 

The  values  of  the  two  components  of  the  velocity  and  of  the  accel- 
eration may  be  expressed  in  terms  of  the  coordinates  of  position  and 
their  time-derivatives. 

272.  Resolution  Parallel  to  Coordinate  Axes. —  If  the  position 
of  the  particle  is  specified  by  its  rectangular  coordinates,  it  is  usually 
convenient  to  resolve  the  displacement,  velocity  and  acceleration 
parallel  to  the  coordinate  axes.     The 

components  in  these  directions  may 
be  called  axial  components. 

273.  Axial  Components  of  Dis- 
placement. —  If  the  particle  moves 
from  A  to  B  (Fig.  1 26)  during  any 
interval  At,  its  total  displacement  is 
given  by  the  vector  AB.  This  total 
displacement  may  be  resolved  into 
components  CD  and  EF,  parallel  to 
the  coordinate  axes.     Let  x,  y  be  the 

coordinates  of  position  at  any  instant ;  let  xx ,  yx  be  the  coordinates 
of  Ay  and  xtJ  y2  those  of  B,  Then  the  axial  components  of  the  dis- 
placement are  evidently  equal  to  x%  —  xx  and  y2  —  yv 

274.  Axial  Components  of  Velocity.  —  The  average  velocity- of 
the  particle  during  the  interval  At  is  (vector  AB)\At ;  the  exact  value 
of  the  velocity  at  the  beginning  of  the  interval  is  the  limit  approached 
by  this  average  value  as  At  is  made  to  approach  zero  (Art.  244). 
But  since  Xt  —  X\  and  y.,  —  yx  are  always  the  values  of  the  axial  com- 


238 


THEORETICAL    MECHANICS. 


ponents  of  vector  AB}  the  axial  components  of  limit  [(vector  AB)/At] 
are 

limit  [(x2  — ^i)/A/]     and     limit  [(j2  — y^/At]. 

The  axial  components  of  the  velocity  are  therefore 

limit  [{xt  —  x^/At]  =  limit  [A*/A/]  =  dx/dt  =  x 
=  ;r- component ; 

limit  [(y.z  —  yx)/At]  =  limit  [Ay  I  At]  =  dyjdt  =  y 
=  j^-component. 

Let  s  denote  the  length  of  the  path  measured  from  some  fixed 
point  up  to  the  position  of  the  particle  ;  v  the  magnitude  of  the  re- 
sultant velocity  ;  a,  /3  the  angles  between  v  and  the  x-  and  j-axes 
respectively.     Then 

V  =  ds/dt ; 

x  =  dxjdt  =  v  cos  a  =  (ds/'dt)  cos  a ; 

y  =  dy/dt  =  v  cos  ft  =  (ds/df)  cos  /3 ; 

tan  a  =  cotan  ft  =  j//ir. 

275.  Axial  Components  of  Velocity-Increment. —  Let  velocities 
be  represented  by  vectors  laid  off  from  a  point  as  in  Art.  245.     Let 

O'  (Fig.  127)  be  this  point,  and  from 
O'  draw  O'X*  and  O'V  parallel  re- 
spectively to  the  axes  of  x  and  y. 
Let  vectors  0'A\  O'B'  represent  the 
values  of  the  velocity  at  the  begin- 
ning and  end  of  an  interval  At.  The 
velocity-increment  for  the  interval  is 
then  represented  by  vector  A'B'. 
This  increment  is  completely  known 
if  we  know  its  axial  components  CD' 
and  E'F'. 

As  found  above,  x  and  y  are  the 
values  of  the  axial  components  of  the 
velocity  at  any  instant.      Let  xx ,  yx  be  the  values  of  these  compo- 
nents at  the  beginning  of  the  interval,  and  xt%  y.,  their  values  at  the 
end  of  the  interval.     Then  (Fig.  127) 

xx  =  0'C\     yx  =  0'E'\     x2  =  0'D\    y2  =  Q'F\ 


PLANE    MOTION    OF    A    PARTICLE. 


239 


Hence  CD'  =  x2  —  xx ;     E'F'  =  y2  — jtx. 

The  axial  components  of  the  velocity-increment  are  therefore 

x2  —  xx  =  Air  =  ^r-component ; 

j.,  —  ji  =  Ay  =  jy-component. 

276.  Axial  Components  of  Acceleration. —  The  average  accel- 
eration for  the  interval  A/  is  equal  to  (vector  A'B') /At.  The  exact 
value  of  the  acceleration  at  the  beginning  of  the  interval  is  the  limit 
approached  by  (vector  A'B')/ At  as  At  is  made  to  approach  zero 
(Art.  250).  Since  the  axial  components  of  vector  A'B'  are  equal  to 
x,  —  xx  and  y2  —  yv  or  Ax  and  Aj>,  whatever  the  length  of  the  in- 
terval, the  axial  components  of  limit  [(vector  A'B')/ At]  are 

limit  [Air/A/]     and     limit  [A>/A/]. 

That  is,  the  axial  components  of  the  acceleration  are 

limit  [Ax  I  At]  =  dx/dt  =  d2x/dt2  =  x  =  ^-component ; 

limit  [Ay /At]  =  dyjdt  =  d2y/dt2  =  y  =  j/-component. 

Let/  denote  the  actual  or  resultant  acceleration  at  any  instant, 
and  a  ,  /3'  the  angles  its  direction  makes  with  the  x-  and  /-axes  re- 
spectively.    Then 

x  =  d'lx\dt2  =  p  cos  a'; 

y  =d2y/dt2  ==/cos/3'; 

p2  =  x2+y2', 
tan  a  =  cotan  /3'  z=y/x. 

277.  Hodograph. —  If  the  vector  representing  the  velocity  at  any 
instant  is  O'P'  (Fig.  128),  O'  being 

a  fixed  point  while  P'  moves  as  the  Yr 
value  of  the  vector  varies,  the  curve  F' 
traced  by  P'  is  the  hodograph  (Art. 

245)- 

The  rectangular  coordinates  of 
the  point  P',  referred  to  axes  O'X', 
O'Y',  parallel  to  the  axes  of  x  and  JE' 
y,  are  x  and  y.  If  the  values  of  x 
and  y  are  known  at  every  instant, 
the  form  of  the  curve  can  be  deter-     O'  C  M'  D'  X1 

mined.  Fig.  128. 


24O  THEORETICAL    MECHANICS. 

278.  Cases  Adapted  to  Polar  Coordinates. — There  are  certain 
problems  to  whose  treatment  polar  coordinates  are  well  adapted. 
This  is  often  the  case  when  the  acceleration  of  the  particle  is  directed 
toward  a  fixed  point.  In  this  case,  and  in  some  others,  it  is  conven- 
ient to  replace  the  acceleration  and  the  velocity  by  their  resolved 
parts  in  directions  parallel  and  perpendicular  to  the  radius  vector. 
The  values  of  these  components  in  terms  of  r  and  6  and  their  deriv- 
atives will  now  be  deduced. 

279.  Components  of  Velocity  Parallel  and  Perpendicular  to 
Radius  Vector. —  Let  a  particle  move  from  A  to  B  along  the  curve 
AB  (Fig.  129)  during  the  interval  At.  Let  r,  6  be  its  polar  coor- 
dinates at  any  time  /,  and  let  the  initial  values  of  these  coordinates 

be  1\ ,  0l ,  and  the  final  values  r% ,  6.2 . 
Then  rx  =  OAtrt  =  OB,  0X  =  A  OXy 
62  =  BOX. 

The  average  velocity  during  the 

interval  is  (vector  AB)/At,  and  the 

true  velocity  at  the  beginning  of  the 

interval  is  the  limit  of  this  value  as 

At  is  made  to  approach  zero.     Take 

Fig.  129.  OA'  =  OA  =  rlt   and  let   (vector 

AB)  be  resolved   into   components 

(vector  A  A')  and  (vector  A' B).     Then  the  true  velocity  at  the 

beginning  of  the  interval  is  equivalent  to  two  components 

limit  [(vector  A  A') /At],     limit  [(vector  A'B)/At]. 
The  latter  component  has  the  direction  OA  and  the  former  is  per- 
pendicular to  OA.      Hence  the  component  of  velocity  parallel  to  the 
radius  vector  is 

limit  [A'B/At]  s=  limit  [(r2  —  rx)/At]  =  limit  [Ar/At]  —  dr/dt, 
and  the  component  of  velocity  perpendicular  to  the  radius  vector  is 
limit  [AA'/At]  =  limit  \rx(0%  —  Oj/At] 

=  limit  \rx  Ad  I  At]  =  r(d6/dt). 
These  results  may  also  be  deduced  from  the  values  of  the  axial 
components  of  velocity.  Thus,  let  x,  y  be  the  coordinates  of  posi- 
tion referred  to  a  pair  of  rectangular  axes,  the  origin  being  at  the  pole 
and  the  ;r-axis  coinciding  with  the  initial  line  of  the  polar  coordinates 
r,  0.     (Fig.  130.)     Then 

x  =  r  cos  6 ;    y  =  r  sin  6.      .         ,         .     (1) 


TH£ 


PLANE  MOTION  OF  A  PARTICLE. 


24I 


The  velocity  is  equivalent  to  a  component  x  in  direction  OX  and  a 
component  y  in  direction  OY.  The  resolved  part  of  the  velocity  in 
any  direction  is  equal  to  the  sum  of 
the  resolved  parts  of  x  and  y  in  that 
direction.  Hence  the  resolved  parts 
of  the  velocity  along  and  perpendic- 
ular to  OP  are  as  follows  : 


Y 

\  Ay    , 

P      &* 

y 

0 

X 

y  sin  6  -f-  x  cos  6  == 

component  along  OP; 

y  cos  6  —  x  sin  6  = 

component  perpendicular  to  OP. 

The  values  of  x  and  y  may  be  ex- 
pressed in  terms  of  r,    6  and  their  derivatives   by   differentiating 
equations  (1)  with  respect  to  /.     Thus, 


Fig.  130. 


dr         a 
x  —  —  cos  v 
dt 

dr 


r  sin  6 


d6 
dt 
d6 


(2) 


y  =  —  sin  0  4-  r  cos  6 

dt  dt 


Substituting  these  values,  the  components  of  velocity  parallel  and 
perpendicular  to  the  radius  vector  become 

dr 
~dt 
dd 


dt 


=  component  in  direction  OP) 

=  component  perpendicular  to  OP. 


280.  Components  of  Acceleration  Parallel  and  Perpendicular 
to  Radius  Vector. —  Since  the  acceleration  is  equivalent  to  two  com- 
ponents, x  in  direction  OX  (Fig. 
131)  and  y  in  direction  OY,  its  re- 
solved parts  parallel  and  perpendicu- 
lar to  OP  are  as  follows  : 


r 

y   / 

P      ?x 

y 

0 

X 

y  sin  6 
y  cos  6 


x  cos  6     in  direction  OP ; 
x  sin  6     perpend,  to  OP. 


Fig.  131, 


The  values  of  x  and  y  in  terms  of  r, 
6  and  their  derivatives  may  be  found 


16 


242  THEORETICAL    MECHANICS. 

by  differentiating  equations  (2),  Art.  279,  with  respect  to  t.  The 
resulting  values  are 

..       ePr        a         drdd   .    a  (d0\2        a  d26  .    a 

x  =  —  cos  u  —  2 sine/  —  r   —     cos  a  —  r sin  u ; 

dt'2  dt  dt  \dt)  dt2 

..       d*r   .    a    .      drdd         a  ldd\2  .  d26        a 

y  = sin  v  -4-  2 cos*,  u  —  r  |  —     sin  u  4-  r cos  v. 

f\        dt2  dtdt  \dt)  dt2 

Hence  .......         Q       d2r  ld0\2 

y  sin  v  4-  x  cos  a  = r\  — 

dt2  \dt) 

=  component  of  acceleration  parallel  to  radius  vector ; 

a        ....  drdO    .      d26 

y  cos  C7  —  .rsin  u  =  2 \-  r  — 

dt  dt  dt2 

=  component  of  acceleration  perpendicular  to  radius  vector. 

The  value  of  the  latter  component  may  be  written  in  another  form, 
as  follows : 

drdd    .      d20       1  dt   td6\ 

2 [-  r = \rl —  . 

dtdt  dt2       rdt\     dt) 

This  may  be  verified  by  differentiation. 

281.  Angular  Motion  of  a  Particle. — The  line  joining  a  mov- 
ing particle  A  to  a.  fixed  point  O  in  general  turns  about  0  as  the 
particle  moves.  Let  6  denote  the  angle  between  OA  and  a  fixed 
line  in  the  plane  of  the  motion  ;  then  the  rate  of  turning  of  the  line  is 
measured  by  the  rate  of  change  of  0.  By  the  ' '  angular  motion  of 
A  about  (9"  is  meant  the  angular  motion  of  the  line  OA. 

Angular  velocity  may  be  defined  as  the  rate  of  angular  motion. 
It  is  measured  by  the  angle  described  per  unit  time.  If  the  line  OA 
turns  at  a  uniform  rate,  let  Ad  be  the  increment  of  6  in  a  time  At, 
and  let  co  denote  the  angular  velocity  ;  then 

co  =  AOjAt. 

If  6  increases  at  a  variable  rate,  AO/At  is  the  average  angular  veloc- 
ity for  the  time  At,  and  the  instantaneous  value  of  the  angular  veloc- 
ity is  the  limit  of  A0/At  as  A^  approaches  o ;  that  is, 

co  =  dOldt. 

A  ngular  acceleration  is  the  rate  of  change  of  angular  velocity. 
If  the  angular  velocity  co  varies  at  a  uniform  rate,  let  Aco  be  its  incre- 


PLANE    MOTION    OF    A    PARTICLE.  243 

ment  during  a  time  At,  and  let  (f)  denote  the  angular  acceleration. 

Then 

<j>  =  Ao>/A/. 

If  co  changes  at  a  variable  rate,  Aco/ A/  is  the  average  angular  accel- 
eration for  the  time  A/,  and  the  instantaneous  value  of  the  angular 
acceleration  is  the  limit  of  Aco/ At  as  At  approaches  o  ;  that  is 

(j>  =  dco/dt. 

If  the  point  A  is  describing  a  circle  of  radius  r  with  center  at  O, 
and  if  As  is  the  length  of  arc  subtended  by  the  angle  A#, 

As = rAO; 
hence 

ds/dt  ==  ridOuit). 

That  is,  the  linear  velocity  is  at  every  instant  equal  to  the  product  of 
the  radius  by  the  angular  velocity.  The  angle  6  must  be  expressed 
in  radians  in  order  that  this  relation  may  hold. 

Examples. 

i.  A  particle  describes  a  circle  in  any  manner.  Show  that  its 
angular  velocity  about  any  point  of  the  circumference  is  at  every  in- 
stant equal  to  half  its  angular  velocity  about  the  center.  The  angular 
velocities  about  all  points  in  the  circumference  are  equal. 

2.  A  particle  P  describes  a  straight  line  at  the  uniform  rate  of  24 
ft.  -per-sec.  Determine  its  angular  velocity  and  angular  acceleration 
about  a  point  A,  6  ft.  from  the  path,  in  two  positions :  (a)  when  AP 
is  at  right  angles  to  the  path,  (b)  when  AP  is  inclined  450  to  the 
path. 

Ans.  (a)  co  =  4  rad. -per-sec,  (f>  =  o.  (b)  co  =  2  rad. -per-sec, 
(f>  —  8  rad. -per-sec. -per-sec. 

3.  A  particle  P  describes  a  straight  line  in  such  a  way  that  its 
angular  velocity  about  a  point  A  distant  h  from  the  path  is  constant 
and  equal  to  00.  Determine  the  linear  velocity  and  linear  accelera- 
tion in  any  position. 

Ans.  Let  6  denote  the  angle  between  AP  and  a  perpendicular  to 
the  path,  v  the  linear  velocity  and  p  the  linear  acceleration  ;  then 
v  =  hco  sec2  6,  p  =  iJico1  tan  6  sec2  6. 

282.  Relation   of  Velocity   and   Acceleration   to   Path. — The 

methods  of  resolution  above  considered,  though  useful  in  the  analy- 
sis of  particular  cases  of  motion,  do  not  show  clearly  the  relation  of 
velocity  and  acceleration  to  the  path  of  the  particle.  This  relation 
is  best  shown  by  resolving  along  the  tangent  and  normal  to  the  path. 


244  THEORETICAL    MECHANICS. 

The  values  of  the  components  of  velocity  and  acceleration  thus  de- 
termined are  independent  of  any  particular  system  of  coordinates  of 
position. 

283.  Tangential  and  Normal  Components  of  Velocity. —  Since 
the  resultant  velocity,  for  any  position  of  the  particle,  has  the  direc- 
tion of  the  tangent  to  the  path,  its  resolved  part  in  the  direction  of 
the  tangent  is  the  velocity  itself,  its  value  being  ds/dt  (Art.  244)  ; 
while  the  resolved  part  along  the  normal  is  zero. 

284.  Tangential  and  Normal  Components  of  Acceleration. — 

Let  A  (Fig.  132)  be  the  position  of  the  particle  at  an  instant  t,  and 

AB  the  path  described  during  an  in- 
terval At.  The  speed  in  the  position 
A  will  be  denoted  by  v. 

From  any  point  O'  lay  off  the 
vector  O'A'  to  represent  the  initial 
velocity  and  O'B'  to  represent  the 
final  velocity  ;  these  vectors  are  then 
parallel  to  the  tangents  to  the  path  at 
A  and  at  B  respectively. 
Fig.  132.  The  velocity-increment  during  the 

interval  is  equal  to  (vector  A'B')  ; 
the  average  acceleration  is  (vector  A' B' )/At;  and  the  true  accelera- 
tion at  the  beginning  of  the  interval  is  the  limit  of  this  average 
value  as  At  is  made  to  approach  zero.  The  vector  A ' B '  will  now 
be  replaced  by  two  components  whose  limiting  directions  are  those 
of  the  tangent  and  normal  to  the  path  at  A.  These  components  are 
A'C,  C'B',  if  the  point  C  is  so  taken  that  O'C  =  0'A'  =  v. 
It  is  evident  that  as  At  approaches  zero,  the  limiting  direction  of 
A'C  is  perpendicular  to  O'A'  and  that  of  C'B'  is  parallel  to  O'A'; 
but  O'A '  is  parallel  to  the  tangent  to  AB  at  A. 

The  tangential  and  normal  components  of  the  acceleration  at  the 
instant  t  are  therefore 

limit  [(vector  C'B')  I  At]     and     limit  [(vector  A'C)  I  At]. 

It  remains  to  determine  the  magnitudes  of  these  limiting  values. 

Evidently  C'B'  is  equal  in  magnitude  to  the  increment  received 
by  the  speed  during  the  interval  At.     Hence 

limit  [C'B' I  At]  =  limit  [Av/At]  =  dv/dt, 


PLANE    MOTION    OF    A    PARTICLE.  245 

which  is  the  value  of  the  tangential  component  of  the  acceleration. 
Next,  to  determine  the  normal  component. 

Draw  normals  to  the  path  at  A  and  B,  and  let  O  be  their  point 
of  intersection.  Lay  off  OC  =  OA.  The  triangles  AOC,  A'O'C 
are  similar ;  hence 

A'C'/A'O'  =  AC/AO; 

AC  =  AC(AO'/AO)  =  AC(v/AO); 

limit  [AC /At]  =  limit  [(v/AO)(AC/At)]. 

As  At  approaches  zero,  the  point  0  approaches  as  a  limiting  position 
the  center  of  curvature  of  the  curve  AB  at  A,  and  the  limiting  value 
of  AO  is  the  radius  of  curvature  at  A.  Call  this  R.  Again,  as  At 
approaches  zero,  the  points  C  and  B  approach  coincidence  ;  A  C  and 
arc  AB  approach  a  ratio  of  equality.      Hence 

limit  [A  CI  At]  =  limit  [(arc  AB)/At]  =  limit  [As /At]  =  ds/dt  =  v. 

Substituting  these  values, 

limit  [A' CI  At]  =  (v/R)v  —  v2/R. 

The  required  components  of  the  resultant  acceleration^  are  there- 
fore 

dvjdt  =  d'2s/dt2  =  tangential  component ; 

v2/R  =  (ds/dt)2/R  =  normal  component. 

These  values  are  independent  of  any  particular  system  of  coordi- 
nates of  position,  but  they  may,  if  desired,  be  expressed  in  terms  of 
the  coordinates. 

Examples. 

1.  Determine  the  magnitude  and  direction  of  the  resultant  accel- 
eration of  a  particle  describing  a  circle  of  radius  r  with  uniform 
speed  v. 

[Express  the  values  of  the  tangential  and  normal  components 
and  determine  their  resultant.  Compare  with  result  of  Ex.  4,  follow- 
ing Art.  251.] 

2.  The  path  of  a  particle  is  a  circle  of  radius  r  lying  in  a  vertical 
plane.  The  speed  is  given  by  the  equation  v  =  k\/z,  where  z  is 
the  vertical  distance  below  a  horizontal  plane  through  the  center  of 
the  circle.  Compute  the  tangential  and  normal  components  of  the 
acceleration  for  any  value  of  z. 

Ans.  Tangential  component  =  (k2 1 2r)\/ (r2 —  z2) ;  normal 
component  —  k2z/r. 


246  THEORETICAL    MECHANICS. 

3.  In  Ex.  2,  let  v  =  8  ft.-per-sec.  when  s  =  1  ft.,  and  let  the 
radius  of  the  circle  be  2  ft.  Compute  the  tangential  and  normal  com- 
ponents of  the  acceleration  when  the  particle  is  in  its  lowest  position. 

Ans.  Tangential  component  —  o;  normal  comoonent  =  64  ft.- 
per-sec-per-sec. 

4.  A  particle  describes  a  circle  01  50  ft.  radius  in  such  a  way  that 
s  =  16/2,  in  which  s  is  in  feet  and  /  in  seconds.  Compare  the  true 
acceleration  when  t  =  2  with  the  average  acceleration  for  the  ensu- 
ing o.  1  second. 

§  2.   Motion  in  a  Plane   Under  Any  Forces. 

285.  Conditions  Under  Which  the  Motion  Will  Be  Confined 
to  a  Plane. —  If  the  direction  of  the  resultant  force  acting  upon  a 
particle  is  always  parallel  to  a  fixed  plane,  and  if  the  motion  at  any 
instant  is  parallel  to  that  plane,  the  resolved  part  of  the  velocity  per- 
pendicular to  the  plane  will  remain  zero,  and  the  path  of  the  particle 
will  therefore  lie  in  the  plane. 

286.  Two  Independent  Equations  of  Motion. —  The  general 
equation  of  motion,  P  =  mp,  is  a  vector  equation,  expressing  equality 
of  direction  as  well  as  of  magnitude  between  the  two  members.  If 
both  Pandfl  be  resolved  in  any  direction,  we  have 

(resolved  part  of  P)  =  m  X  (resolved  part  of  f). 

Hence  by  resolving  in  different  directions,  any  number  of  true  equa- 
tions may  be  written.  In  the  case  of  plane  motion,  however,  only 
two  of  these  equations  can  be  independent.  For  if  the  resolved  parts 
of  a  vector  in  two  directions  are  given,  the  vector  is  completely  de- 
termined ;  hence  the  two  equations  obtained  by  resolving  in  two  dif- 
ferent directions  include  all  that  is  expressed  by  any  equation  ob- 
tained by  resolving  in  a  third  direction. 

The  form  of  the  two  independent  equations  depends  upon  the 
system  of  coordinates  employed  in  specifying  the  position  of  the  par- 
ticle, and  also  upon  the  directions  of  resolution. 

287.  Equations  of  Motion  in  Terms  of  Rectangular  Coordi- 
nates.—  Let  the  position  of  a  particle  be  specified  by  its  coordinates 
x  and  j/,  referred  to  any  pair  of  fixed  rectangular  axes  in  the  plane  of 
the  motion,  and  let  the  two  equations  of  motion  be  obtained  by  re- 
solving parallel  to  the  axes.  Let  the  axial  components  of  the  result- 
ant force  P  be  X  and  Y.      If  several  forces  act  upon  the  particle,  X 


PLANE    MOTION    OF    A    PARTICLE.  247 

is  the  algebraic  sum  of  their  resolved  parts  in  the  ^-direction,  and  Y 
the  algebraic  sum  of  their  resolved  parts  in  the  ^-direction.  The 
axial  components  of  the  acceleration  are  x  and  y  (Art.  276).  The 
equations  of  motion  are  therefore 

X  =  m'x  =  m(d%xldt*) ;  .         .     (1) 

Y  =  my  ==  m(d2yldt2).  .  .         .     (2) 

288.  Equations  of  Motion  in  Terms  of   Polar  Coordinates. — 

If  polar  coordinates  are  employed,  the  equations  are  obtained  in  the 
most  convenient  form  by  resolving  parallel  and  perpendicular  to  the 
radius  vector.  Let  Pr  and  Pg  be  the  resolved  parts  of  the  resultant 
force  P  in  these  directions  respectively.  The  values  of  the  resolved 
parts  of  the  acceleration  are  given  in  Art.  280.  Using  these  values, 
the  equations  of  motion  become 


Pr  ---=  m 


■<Pr_   M\n 
dt2      \dt) 


(  drdd    ,      d26\  ri  dt  2d6 

2 (-  r —    =  m\ [r2  — 

\   dtdt  dt2}  \rdt\     dt 


■     (1) 


(2) 


289.  Resolution  Along  Tangent  and  Normal  to   Path. —  Let 

the  components  of  the  resultant  force  P  resolved  along  the  tangent 
and  normal  to  the  path  at  the  instantaneous  position  of  the  particle 
be  denoted  by  Pt  and  Pn  respectively.  Using  the  values  of  the  re- 
solved parts  of  the  acceleration  in  these  directions  (Art.  284),  the 
equations  of  motion  become 

Pt  =  midvldt)  =  m(d2s/dt2)  ;  .         .     (1) 

Pn  =  m(v2/R) (2) 

290.  Classes  of  Problems. —  Problems  relating  to  the  motion  of 
a  particle  in  a  plane  may  be  classed  as  inverse  or  direct,  according  as 
their  solution  does  or  does  not  involve  integration.  (See  Art.  223.) 
The  most  important  cases  fall  under  one  of  the  following  general 
problems : 

(a)  To  determine  the  resultant  force  at  any  instant,  the  motion 
being  completely  known. 

(b)  To  determine  the  motion  when  the  forces  are  known. 
The  former  problem  is  direct,  the  latter  inverse. 


248  THEORETICAL    MECHANICS. 

291.  Direct  Problem:  To  Determine  the  Resultant  Force  When 
the  Motion  Is  Known.  —  If  the  acceleration  is  known,  the  resultant 
force  is  given  immediately  by  the  general  vector  equation  P  =  mfl, 
or  by  any  pair  of  algebraic  equations  equivalent  to  it.  If  the  coor- 
dinates of  position,  or  the  components  of  the  velocity,  are  known 
functions  of  the  time,  the  components  of  the  acceleration  can  be  de- 
termined by  differentiation. 

Examples. 

1 .  A  body  describes  any  curve  at  a  uniform  speed  ;  determine 
the  resultant  force  acting  upon  it. 

2.  A  body  of  mass  12  lbs.  describes  the  circumference  of  a  circle 
12  ft.  in  diameter  at  the  uniform  rate  of  100  ft.-per-min.  Required 
the  resultant  force  acting  upon  it. 

3.  The  position  of  a  particle  at  any  time  is  given  by  the  equations 
x  =  a  -f-  bt^y  y  =  A  -\-  Bf.  The  mass  being  m,  determine  the 
magnitude  and  direction  of  the  resultant  force  at  the  time  /. 

4.  Determine  the  magnitude  and  direction  of  the  resultant  force 
at  time  t  if  the  position  is  given  by  the  equations  x  =  a  +  btt  y  = 
A+Bt+  Ct\ 

5.  What  is  the  resultant  force  acting  on  the  particle  P  in  Ex.  3, 
Art.  281?  If  h  =  6  ft.,  co  =  1  rad. -per-sec. ,  and  the  mass  of  the 
particle  is  5  lbs.,  determine  the  value  of  the  resultant  force  when  the 
particle  is  12  ft.  from  A.  Ans.  415.7  poundals. 

6.  The  path  of  a  particle  is  an  ellipse  whose  semi-axes  are  20  ft. 
and  10  ft.  The  mass  is  12  lbs.  and  the  speed  is  uniformly  100  ft.- 
per-min.  Determine  the  resultant  force  acting  on  the  particle  when 
at  the  extremity  of  each  major  axis. 

292.  Inverse  Problem. —  If  the  forces  are  known,  the  complete 
determination  of  the  motion  requires  the  integration  of  two  simulta- 
neous differential  equations.  These  may  be  of  various  forms,  de- 
pending upon  the  system  of  coordinates  employed  and  also  upon  the 
directions  of  resolution.  Usually  it  will  be  found  convenient  to  use 
one  of  the  three  pairs  of  equations  given  in  Arts.  287,  288,  289. 

A  problem  of  this  class  cannot  be  completely  solved  unless  suf- 
ficient initial  conditions  are  given  for  the  determination  of  the  con- 
stants of  integration. 

We  proceed  to  the  consideration  of  some  important  cases  of  plane 
motion. 


PLANE    MOTION   OF    A    PARTICLE.  249 

§  3.  Resultant  Force  Constant  or  Zero. 

293.  Direction  of  Resultant  Force  Constant. —  If  the  resultant 
force  has  a  constant  direction,  let  this  be  taken  as  the  direction  of 
the  axis  of  x ;  then  X  =  P,  Y  =  o ;  and  the  equations  of  motion 

(Art.  287)  are 

m(d2x/dt2)  =  P;       .         .         .         .     (1) 

m(d2y!dt2)  == o (2) 

Equation  (1)  cannot  be  integrated  unless  Pis  given:      From  (2), 

dyjdt  =  C  =  constant ;  (3) 

y=Ct+  C (4) 

Equation  (3)  expresses  the  fact  that  the  ^/-component  of  the  velocity 
remains  constant.      Motion  of  this  character  is  often  called  parabolic. 
The  values  of  the  constants  C  and  C  may  be  determined  if  the 
values  of y  and  dyjdt  at  some  instant  are  known. 

294.  Resultant  Force  Constant  in  Direction  and  Magnitude. — 

If  the  resultant  force  is  constant  in  magnitude  as  well  as  in  direction, 
equation  (1)  can  be  integrated  directly.      Let 

Pjm  =f=  constant  ; 

then  d2x/dt'2=f. 

Integrating  twice, 

d*/dt=/t+  a  .     .     .     .  (5) 

x  =  ift2  -f  C"t  +  C".        .        .        .     (6) 

Equations  (4)  and  (6)  give  the  position  at  any  time,  and  (3)  and  (5) 
the  velocity  at  any  time,  as  soon  as  the  constants  C,  C ,  £"'and  C" 
are  known.  These  may  be  found  if  the  position  and  velocity  at  one 
instant  are  known. 

295.  Projectile. —  A  particle  projected  in  any  direction  and  left 
to  the  action  of  gravity  and  the  resistance  of  the  air  is  called  a 
projectile.  In  the  following  discussion  the  resistance  of  the  air  is 
disregarded. 

This  problem  is  a  special  case  of  that  treated  in  Art.  294.  The 
solution  there  given  applies,  if  the  axes  of  coordinates  are  properly 
chosen.     Obviously  the  axis  of  x  must  be  vertical  and  the  axis  of  y 


250 


THEORETICAL    MECHANICS. 


X' 

Fig.  133. 


horizontal,  and g  must  replace/".      In  order  to  determine  the  con- 
stants of  integration,  let  the  following  conditions  be  specified  : 

Let  the  particle  be  projected  from  , 
a  certain  point  which  will  be  taken  as 
origin  of  coordinates,  the  positive  di- 
rection for  x  being  downward.  Let 
the  velocity  of  projection  be  V,  its  di- 
rection making  an  angle  a  upward 
from  the  plus  direction  of  the  j-axis. 
Let  /  be  reckoned  from  the  instant 
when  the  particle  is  at  the  origin  O 

(Fig-  133). 

The  equations  expressing  the  val- 
ues of  the  axial   components  of  the 
velocity  and  of  the  coordinates  of  position  are,  as  above  found, 

dxjdt  =gt  +  C'\     dyjdt  =  C\ 

x  =  igt2  +  C"t  +  C")    y=Ct  +  C. 
The  initial  conditions  are  that  when  /  =  o, 

x  =  o,    y  =  o,     dxjdt  =  —  V  sin  a,     dyjdt  ==   V  cos  a. 
The  values  of  the  constants  are  therefore 

C  =   V  cos  a,      C  =  0,      C"  =  —  V  sin  a,      C"  =  o, 
and  the  equations  become 

dxjdt  =  gt —   Fsina;  .  .  .      (1) 

dyjdt  =  V  cos  a ;         .         .         .         .     (2) 

*=\gt2—  Fsina-/;  .  .      (3) 

y  =  V  cos  a  -  t.  .  .  •      (4) 

Equations  (1)  and  (2)  give  the  velocity  at  any  time,  and  equations 
(3)  and  (4)  the  position. 

Equation  of  path. —  Eliminating  /  between  (3)  and  (4),  the  equa- 
tion of  the  path  is  found  to  be 


2  V'2  cos2  a 


—  y  tan  a. 


(5) 


This  is  the  equation  of  a  parabola  with  its  axis  vertical. 

The  position  of  the  vertex  of  the  parabola  may  be  found  by  so 
transforming  coordinates  that  the  equation  takes  the  form  yl  =  \mx. 


PLANE    MOTION    OF    A    PARTICLE.  251 

If  we  put  dx/dt  =  o  in  (i)  we  find  t  =  (  Fsin  a)/g;  this  value  of  / 
substituted  in  (3)  and  (4)  gives 

x  =  —  (  V2  sin2  d)\2g ;    y  =  (  V2  sin  a  cos  a)/g. 

These  are  the  coordinates  of  the  point  at  which  the  curve  is  horizon- 
tal, and  this  is  the  vertex  of  the  parabola.  For  if,  with  this  point  as 
origin,  the  coordinates  of  any  point  of  the  curve  are  x\  y\  we  have 

X  =  x'  —  (  V2  sin2  a)/2g,     y  =  y'  +  (  V2  sin  a  cos  a)/g. 

These  values  substituted  in  (5)  give 

,,        2  V2  cos2  a     , 

y   = x\ 

g 

which  is  the  equation  of  a  parabola  whose  vertex  is  at  the  origin 
O'  and  whose  principal  diameter  is  in  the  axis  of  x' . 

Examples. 

1.  Determine  the  distance  from  O  (Fig.  133)  at  which  the  body 
will  cross  the  line  0  Y.  (This  distance  is  called  the  range  of  the 
projectile.)  Ans.  (  V2  sin  2a)/g. 

2.  Determine  for  what  value  of  a  the  range  is  greatest. 

Ans.  a  =  450. 

3.  Determine  the  greatest  height  to  which  the  projectile  will  rise. 

Ans.  (  V2  sin2  a)J2g. 

4.  If  the  body  is  projected  with  a  velocity  of  50  ft.-per-sec.  in  a 
direction  inclined  25°  upward  from  the  horizontal,  find  (a)  the  equa- 
tion of  the  path  ;  (b)  the  position  of  the  highest  point  reached  with 
reference  to  the  point  of  projection  ;  (c)  the  range  ;  and  (d)  disposi- 
tion and  velocity  at  the  end  of  4  sec.  from  the  instant  of  projection. 

5.  Prove  that  the  same  range  will  result  from  two  different  values 
of  a.      How  are  these  two  values  related  ? 

An s.  They  are  complementary. 

6.  A  particle  is  projected  from  a  point  A  with  a  velocity  V.  What 
must  be  the  direction  of  projection  in  order  that  it  may  pass  through 
a  point  B  such  that  the  line  AB  is  inclined  at  angle  6  to  the  horizon 
and  the  distance  AB  =  a? 

Ans.  If  a  =  angle  between  V  and  horizontal,  its  value  is  given 
by  the  equation 

sin  (2a  —  0)  =  sin  6  +  (ga  cos2  6)1  V2. 
Show  that  the  problem  is  impossible  unless  V2  is  at  least  as  great  as 
ga{\  +  sin  (9). 

7.  A  particle  is  to  be  projected  from  a  given  point  in  such  a  way 
as  to  pass  through  a  point  50  ft.  higher  than  the  point  of  projection 


252  THEORETICAL    MECHANICS. 

and  200  ft.  distant  from  it.    What  is  the  least  allowable  speed  of  pro- 
jection ?    What  is  the  corresponding  direction  of  projection  ? 

Ans.    V=  89.7  ft. -per-sec. ;  a  =  520  14'. 

296.  Resultant  Force  Zero. —  If  the  resultant  of  all  forces  acting 
upon  a  particle  is  zero  at  any  instant,  the  acceleration  is  also  equal  to 
zero.  If  the  resultant  force  remains  zero  during  any  interval,  the 
acceleration  remains  zero  throughout  that  interval.  The  velocity 
therefore  remains  constant  in  magnitude  and  direction.* 

297.  Statics  a  Special  Case  of  Kinetics. —  Statics  has  been  de- 
fined as  that  branch  of  Dynamics  which  treats  of  the  conditions  of 
equivalence  of  systems  of  forces,  and  especially  of  the  conditions 
under  which  forces  balance  each  other  or  are  in  equilibrium.  (Art. 
9.)  The  principles  of  Statics,  although  they  may  be  developed  to  a 
great  extent  independently  of  the  laws  of  motion  (as  shown  in  Part 
I),  are  also  included  in  the  principles  of  Kinetics. 

Thus,  in  developing  the  laws  of  motion  we  have  made  use  of  the 
principle  that  the  resultant  of  any  system  of  concurrent  forces  is  a 
force  equal  to  their  vector  sum.  The  same  principle  was  used  in  deal- 
ing with  concurrent  forces  in  Part  I.  All  the  methods  of  combining 
and  resolving  concurrent  forces  which  have  been  developed  in  Statics 
are  therefore  included  in  Kinetics. 

It  will  be  noticed  that  the  principle  of  the  parallelogram  (or  tri- 
angle) of  forces,  which  is  the  foundation  of  the  methods  of  combining 
concurrent  forces,  has  not  been  proved,  but  has  been  assumed  as  a 
fundamental  axiom,  both  in  the  development  of  Statics  (Art.  59)  and 
in  that  of  Kinetics  (Art.  255). 

298.  Equilibrium. —  A  system  of  forces  applied  to  a  particle  is 
in  equilibrium  if  their  combined  action  produces  no  effect  upon  the 
motion  of  the  particle.  The  general  condition  of  equilibrium  for 
any  set  of  forces  is,  therefore,  that  their  resultant  is  zero.  This  does 
not  imply  that  the  particle  is  at  rest  ;  but  that  its  acceleration  is  zero 
and  its  velocity  uniform  in  magnitude  and  direction.  This  result  is 
included  in  the  meaning  of  the  general  equation  of  motion  (Art. 
256),  as  of  course  it  should  be. 

*  This  is  not  given  as  a  proof  of  the  proposition  that  a  particle  acted 
upon  by  no  forces,  or  by  forces  whose  resultant  is  zero,  moves  in  a  straight 
line  at  a  uniform  rate.  This  principle  has  been  assumed  in  the  deduction  of 
the  general  equation  of  motion,  which  must  of  course  include  the  principle 
itself. 


PLANE    MOTION   OF    A    PARTICLE.  253 

§  4.    Central  Force. 

299.  Equations  of  Motion  of  Particle  Acted  Upon  by  Central 
Force. —  A  force  which  is  always  directed  toward  or  from  a  fixed 
point  is  called  a  central  force.  The  fixed  point  is  a  center  of  attrac- 
tion or  a  center  of  repulsion  according  as  the  force  is  directed  toward 
or  from  the  center. 

In  many  cases  polar  coordinates  will  be  found  most  convenient  in 
discussing  the  motion  of  a  particle  under  a  central  force.  We  give 
the  form  taken  by  the  differential  equations,  using  both  rectangular 
and  polar  coordinates. 

Rectangular  coordinates. —  Let  the  center  of  force  be  taken  as 
origin  of  coordinates,  and  let  N  (Fig.  134)  be  the  position  of  the 
particle  at  the  time  /.  If  the  resultant  force  is  an  attraction  of  mag- 
nitude P,  we  have 

X  =  —  P  cos  0  =  —Pix/r)  ; 

Y=  —Psm  0  ==  —  P(y/r). 

The  equations  of  motion  are  therefore 

m(d2x/dt2)  = —P(x/r);        .         .         .     (i) 

m(dy/dt*)  =  -P(y/r);       .         .         .     (2) 

in  which  r  =  v  x2  -j-  y'1. 

Polar  coordinates.  —  Take  the  center  of  force  as  pole,  and  choose 
any  initial  line.  Resolving  forces  and  accelerations  parallel  and  per- 
pendicular to  the  radius  vector,  the  equations  of  motion  given  in  Art. 
288  become 

dt2  \dt)  m  ^ 


id_[^dO\ 
rdt 


(4:h-    ■  •  • « 


In  the  important  case  treated  in  the  next  Article  it  is  convenient 
to  employ  rectangular  coordinates.  The  subsequent  discussion  of 
central  forces  will  refer  mainly  to  the  equations  in  polar  coordinates 
and  their  application  to  particular  cases. 

300.  Force  Varying  Directly  as  the  Distance  From  a  Fixed 
Point. — When  the  central  force  Varies  directly  as  the  distance  from 


254 


THEORETICAL    MECHANICS. 


the  fixed  point,  the  equations  in  rectangular   coordinates  are  readily 
integrated. 

Let  k  denote  the  attraction  per  unit  mass  at  unit  distance  from 
the  center;  then  P/m  =  kr,  and  equations  (i)  and  (2)  of  Art.  299 
become 

d*x(dt*  =  —kx)       .        .        .        .     (5) 

d2y/dt2  =  —ky (6) 

Each  of  these  equations  is  identical  in  form  with  equation  (1)  of  Art. 

229,  and  they  may  be  integrated  sepa- 
rately by  the  method  there  employed. 
Four  constants  of  integration  will  be  in- 
troduced. In  order  to  determine  their 
values,  let  the  following  initial  condi- 
tions be  assumed  : 

At  a  certain  instant  the  particle  is  in 

the  axis  of  x  at  a  distance  a  from  the 

origin,    and   is   moving   parallel   to   the 

axis   of  y  with  velocity    V.      Let   t  be 

Then   the  initial   conditions   may  be 


M  X 


Fig.  134. 


reckoned  from   this   instant 
stated  as  follows  : 

When  t  =  o,  x  =  a,  y  =  o,  x  =  o,  y 


V. 


Integrating  equation  (5)  as  in  Art.  229,  and  determining  the  con- 
stant from  the  condition  that  x  =  o  when  x  =  a,  there  results 


(dx/dty  =  k(a2  —x2) 


(7) 


Integrating  again  (Art.  229),  and  determining  the  constant  from  the 
condition  that  x  =  a  when  /  =  o, 

kHt  =  s'm-^x/a)  —  sin_1(i). 

Taking  cosine  of  each   member,  and  multiplying  by  a,  the  result  is 

x  =  a  cos  (k*f) (8) 

Integrating  equation  (6)  and  determining  the   constant   by  the 
condition  that  y  =   V  when  y  =  o, 

yl  =  (dyldff  =  V2  —  ky2  =  k{b2  —  y2\     .         .     (9) 

if  b2  is  written  for  V2jk.     A  second  integration  gives 

kHt  --=  &xrx(ylb)  —  sin-'Co), 


PLANE    MOTION    OF    A   PARTICLE.  255 

the  constant  being  determined  by  the  condition  that  y  ==  o  when 
t  =  o.     Taking  sine  of  each  member, 

sin  (kHt)  =  ±  y/b. 

The  +  sign  must  be  used  if  the  particle  is  moving  in  the  positive 
j-direction  when  t  =  o,  because  it  is  evident  that  y  is  +  for  small 
values  of  /.     The  equation  may  therefore  be  written 

y  =  b  sin  (k*t) (10) 

Equations  (8)  and  (10)  give  the  position,  and  (7)  and  (9)  the 
velocity,  at  every  instant. 

Repulsive  force. —  If  the  force  is  repulsive,  the  equations  of  mo- 
tion must  be  changed  by  the  substitution  of  a  minus  quantity  for  k. 
The  above  solution  holds  mathematically  for  this  case,  but  involves 
imaginaries.  This  may  be  avoided  by  integrating  the  equations  by 
the  method  employed  in  Art.  230.  If  initial  conditions  are  assumed 
as  in  the  above  solution  of  the  case  of  attraction,  the  values  of  x  and 

Jare  x=Mek* '  +  *-**');      .        •        .    (11) 

y  =  id(e**' —  e-**').        .         .         .     (12) 

In  these  equations  k  means  the  repulsive  force  per  unit  mass  at  unit 
distance  from  the  center,  and  b  =   V[\/k. 

Examples. 

1.  Show  by  eliminating  t  between  equations  (8)  and  (10)  that 
the  path  is  an  ellipse  whose  principal  diameters  lie  in  the  coordinate 
axes.      Determine  the  values  of  the  principal  semi-diameters. 

2.  Determine  the  time  of  describing  the  whole  ellipse.  Prove 
that  this  depends  upon  the  intensity  of  the  force  (i.  e. ,  its  value  at  a 
given  distance  from  the  center),  and  is  independent  of  the  initial  con- 
ditions. 

3.  What  are  the  dimensions  of  the  constant  /£? 

4.  In  case  of  a  repulsive  force,  show  that  the  path  is  an  hyper- 
bola.     Determine  its  equation. 

301.  Central  Force  Varying  According  to  Any  Law. —  Re- 
turning to  the  general  case  of  motion  under  a  force  directed  toward 
a  fixed  point,  certain  general  principles  may  be  deduced  which  hold 
without  restricting  the  law  of  variation  of  the  force. 

Using  polar  coordinates  and  resolving  along  and  perpendicular  to 
the  radius  vector,  the  equations  of  motion  are  (Art.  299) 


256  THEORETICAL    MECHANICS. 

■(£)'= --;    •  •  •  (i> 

\dti             m 
(r^Uo (2) 


d2r  idO\ 

dt2 


ld_ltde\ 

rdr 


Here  P  is  the  magnitude  of  the  force.  To  determine  the  motion 
these  equations  must  be  integrated.  For  the  complete  integration, 
P  must  be  a  known  function  of  one  or  more  of  the  variables  r,  6,  t. 
In  the  most  important  case  P  is  a  function  of  r  only.  One  important 
result  may,  however,  be  obtained  for  any  case  of  motion  under  a 
central  force. 

From  equation  (2),  multiplying  by  r  and  integrating,  we  have 

J6        7 
r  —  =  h  =  constant.  .         .  (i) 

dt 

To  interpret  this  result,  let  A  denote  the  area  swept  over  by  the 
radius  vector,  in  moving  from  some  given  position.  Thus,  in  Fig. 
1 34,  let  N  be  the  position  of  the  particle  at  time  t,  ON  =  r, 
NOX  =  6 ;  and  let  MN  be  a  portion  of  the  path,  described  in  the 
direction  MN.     Then 

A  =  area  MON 

In  time  At  let  the  particle  move  to  TV';  the  increments  of  6  and  A 

are 

Ad  =  angle  NON\     AA  =  area  NON'. 

Approximately, 

AA  =  ir2A0,     or    AA/At  =  lr\A6/At). 

In  the  limit,  as  At  is  made  to  approach  zero,  the  equation  is  exact ; 
that  is, 

r\d6/dt)  =  2(dA/dt). 

Equation  (3)  may  therefore  be  written 

dA/dt  =  kJ2  =  constant.         .         .         .     (4) 

The  quantity  dA/dt  may  be  called  the  areal  velocity  of  the  par- 
ticle, being  the  rate  at  which  the  radius  vector  is  describing  area  at 
any  instant.  The  last  equation  therefore  shows  that  the  areal  veloc- 
ity is  constant ;  and  this  conclusion  evidently  holds  for  all  cases  of 
motion  of  a  particle  in  a  plane,  so  long  as  the  acceleration  is  directed 
toward  a  fixed  point, 


PLANE    MOTION    OF    A    PARTICLE.  257 

Integrating  equation  (4)  between  limits  corresponding  to  instants 
tx  and  t2 , 

A2-A,  =  \h(t.t  -  O- 

That  is,  the  area  described  by  the  radius  vector  during  any  interval 
is  proportional  directly  to  the  interval. 

The  constant  h  is  the  double  areal  velocity.  Its  value  is  known 
in  any  case  if  the  velocity  in  some  one  position  is  known  in  magni- 
tude and  direction.  Thus,  at  a  certain  instant  let  the  distance  from 
the  center  of  force  be  a,  and  let  the  resolved  part  of  the  velocity  at 
right  angles  to  the  radius  vector  be  V.  The  areal  velocity  at  that 
instant  is  a  V/2,  and  therefore  h  =  a  V. 

As  applied  to  the  motion  of  a  planet  under  the  attraction  of  the 
sun,  equation  (4)  expresses  Kepler's  law  that  the  radius  vector  de- 
scribes equal  areas  in  equal  times.  The  law  is  seen  to  be  true,  not 
only  when  the  force  follows  the  law  of  gravitation,  but  in  every  case 
of  central  force. 

For  use  in  the  treatment  of  certain  problems,  equation  (i)  takes 
a  more  convenient  form  if  a  new  variable,  u  =  i/r,  be  introduced 
instead  of  r.     This  is  accomplished  as  follows  : 
From  equation  (3) 

5  =  ^! (5) 

at 

Also,  differentiating  the  equation  r  =  i/«, 

dr  _         1  du  _         1  dudB  _        ,du 

dt~         u2dt~         u2d6~dt~  dO 

(putting  for  dd/dt  its  value  hu2).     Again, 

d2r  _    _,dldu\_  d2udd  _  2d2u 

dt2~         dt\d6!~    ~    a7e2a7t~    ~    *  dO2' 

Substituting  the  values  of  dO/dt  and  d2r/dt2  in  (1),  we  have  finally 

P  d2u 

-  =  h2u2—  +  h2u\  .         .         .     (6) 

m  dd2 

Equations  (5)  and  (6)  may  be  used  instead  of  (1)  and  (2). 

To  determine  the  law  of  variation  of  the  force  zvhen  the  path  is 
given. — If  the  polar  equation  of  the  path  is  given,  d2ujd62  may  be 
found  by  differentiation.      By  substituting  its  value  in  equation  (6), 
the  force  may  be  determined  as  a  function  of  0>  u  or  r. 
17 


258  theoretical  mechanics. 

Examples. 

1 .  Show  that  a  particle  cannot  describe  a  straight  line  under  a 
force  directed  toward  or  from  a  fixed  point  not  in  the  path. 

[Applying  the  above  method  of  determining  the  law  of  force,  it 
will  appear  that  P  =  o.  ] 

2.  If  the  orbit  is  a  circle  passing  through  the  center  of  force, 
show  that  the  force  varies  inversely  as  the  fifth  power  of  the  distance. 

3.  A  particle  describes  a  conic  section  under  the  action  of  a  force 
which  is  always  directed  toward  a  focus.  Show  that  the  force  varies 
inversely  as  the  square  of  the  distance. 

302.  Force  Varying  Inversely  as  the  Square  of  the  Distance. — 

The  most  important  case  of  motion  under  a  central  force  is  that  in 
which  the  force  varies  inversely  as  the  square  of  the  distance  from 
the  fixed  point.  In  this  case  it  is  most  convenient  to  employ  equa- 
tions (5)  and  (6)  of  Art.  301.  Let  the  attraction  per  unit  mass  at 
unit  distance  from  the  center  be  k  ;  then  P/m  =  k/r'1  =  ku1,  and 
equation  (6)  reduces  to  the  form 

k        dlu    . 

+  * (7) 


This,  and  the  equation 


h1       dO'1 

$  =  **«,  ....      (8) 

at 


are  equivalent,  in  the  present  case,  to  the  general  differential  equa- 
tions of  motion. 

The  path  of  the  particle  may  be  found  by  the  integration  of 
equation  (7).  To  accomplish  this,  put  u  =  z  -f-  k//i2,  and  there 
results 

(Pz____ 

d6*  ~ 

Proceeding  as  in  Art.  229,  the  first  integration  gives 

S)=-*!+c'    •  •  • (9) 

and  the  second, 

z  =  C  sin  (Q  4  «)• 

Here  C  and  a  are  constants  whose  values  must  be  found  from  the 
initial  conditions.      Restoring  the  value  of  gt 

k 
u== 1-  Csin  (0  +  a).  .  .      (10) 


PLANE    MOTION    OF    A    PARTICLE.  259 

The  form  of  this  equation  shows  that  the  curve  is  a  conic  section, 
the  focus  being  at  the  pole  or  center  of  force.  To  simplify  its  form, 
let  C  and  a  be  determined  from  the  condition  that  when  6  =  0,  r  is 
equal  to  a,  and  the  velocity  is  perpendicular  to  the  radius  vector. 
The  last  condition  makes  drjdt,  dujdt  and  dnjdQ  all  zero  when 
0  =  o.     From  (10), 

Hence  the  condition  dujdO  =  o  when  6  =  o  gives 

cos  a  =  o ;     sin  a  =  1 . 
Also  from  (10),  applying  the  condition  that  u  •=  i/a  when  6  =  o, 

^. "  I     _  k 

The  equation  of  the  path  is  therefore 

h1       \a       /W  lil\_  \ak  I  J 

This  represents  a  conic  section  of  eccentricity  k2/a&  —  1  and  semi- 
parameter  k2/k,  the  focus  being  at  the  pole  of  the  system  of  coor- 
dinates and  the  principal  diameter  lying  in  the  initial  line. 

Since  a  conic  section  is  an  ellipse,  a  parabola,  or  an  hyperbola, 
according  as  the  eccentricity  is  less  than,  equal  to,  or  greater  than 
unity,  it  is  seen  that  the  orbit  is 

an  ellipse  if    Jffak  <  2  ; 
a  parabola  if    Hl\ak  =  2  ; 
an  hyperbola  if    h2/ak  >  2. 

Since  h  depends  upon  the  velocity  in  the  initial  position,  it  is  seen 
that  if  the  initial  velocity  has  a  certain  value  the  orbit  is  a  parabola ; 
if  it  is  less  than  this  value  the  orbit  is  an  ellipse ;  and  if  greater,  the 
orbit  is  an  hyperbola.  To  find  this  critical  velocity,  let  V  denote  the 
velocity  when  r  =  a  and  6  =  o,  its  direction  being  perpendicular  to 
the  radius  vector.  The  double  areal  velocity  k  has  the  value  a  V ; 
hence  hl\ak  =  a  V'zlky  and  the  above  relations  become : 

V'2  <  2k/a     for  ellipse  ; 
V2  =  ik\a     for  parabola  ; 
V1  >  ik\a     for  hyperbola. 


260  theoretical  mechanics. 

Examples. 

i  .  A  body  moves  under  the  action  of  a  central  force  which  varies 
inversely  as  the  square  of  the  distance.  At  a  certain  instant  it  is  10 
ft.  from  the  center  of  attraction  and  moving  at  right  angles  to  the 
radius  vector  at  the  rate  of  5  ft.-per-sec.  In  this  position  the  force 
is  10  poundals  per  pound  of  mass.  Write  the  equation  of  the  path, 
and  determine  whether  it  is  an  ellipse,  a  parabola  or  an  hyperbola. 

2.  If  v  is  the  velocity  of  the  particle  when  at  a  distance  r  from 
the  center  of  attraction,  show  that  the  path  is  an  ellipse,  a  parabola 
or  an  hyperbola,  according  as  v1  is  less  than,  equal  to,  or  greater  than 
2k\r ;  this  result  being  independent  of  the  direction  of  v. 

3.  Prove  that  v2  —  2k/r  is  constant  when  the  force  varies  in- 
versely as  the  square  of  the  distance. 

4.  Integrate  equation  (6),  Art.  301,  for  the  case  in  which  the 
force  varies  inversely  as  the  cube  of  the  distance. 


§  5.   Constrained  Motion. 

303.  Meaning  of  Constrained  Motion. —  The  motion  of  a  body 
is  said  to  be  constrained  if  certain  conditions  are  imposed  upon  it, 
while  the  forces  which  must  act  in  order  that  the  motion  may  con- 
form to  those  conditions  are  not  specified. 

A  bead  sliding  on  a  wire  of  any  form  which  is  either  at  rest  or 
moves  in  a  specified  manner  furnishes  an  example  of  constrained 
motion.  Another  case  is  that  of  a  body  sliding  on  an  inclined  plane 
which  is  either  at  rest  or  is  moving  in  a  given  manner. 

The  laws  of  motion  and  the  general  equations  deduced  from 
them,  in  any  of  the  forms  given  in  the  foregoing  discussions,  may  be 
applied  in  the  solution  of  problems  in  constrained  motion.  The  con- 
straint is  always  produced  by  the  action  of  forces.  The  feature  which 
distinguishes  constrained  motion  is  that  certain  of  the  forces  are  not 
given  directly,  but  are  given  indirectly  by  specifying  their  effect  upon 
the  motion.     Such  forces  are  called  constraining  forces. 

304.  General  Method. — In  dealing  with  a  case  of  constrained 
motion  of  a  particle,  we  are  to  apply  the  general  differential  equa- 
tions of  motion,  the  constraining  forces  being  introduced  as  unknown 
quantities.  In  addition  there  will  be  one  or  more  equations  express- 
ing the  conditions  which  are  to  be  imposed  upon  the  motion.  If 
rectangular  coordinates  are  used,  we  have  (a) 


PLANE    MOTION    OF    A    PARTICLE.  26 1 

m'x  =  X,     .  .  .  .  .       (i) 

my  =   Y,     .  .  .  .  .       (2) 

in  which  X  and  Y  involve  the  unknown  constraining  forces  as  well 
as  the  known  forces  ;  and  (<£)  certain  equations  of  condition  which 
may  involve  any  or  all  of  the  quantities  x,  y,  t,  x,  y,  x,  y. 

The  only  case  which  will  here  be  considered  is  that  in  which  the 
path  is  assigned.     The  equation  of  condition  is  then  of  the  form 

A*>  y)  =  ° (3) 

Before  discussing  this  problem  in  its  general  form,  the  simple  case  of 
motion  in  a  straight  line  will  be  considered. 

305.  Motion  on  a  Smooth   Inclined   Plane  Under   Gravity. — 
Let  a  body  of  mass  m  pounds  be  placed  upon  a  smooth  plane  sur- 
face, inclined  to  the  horizontal  at  an  angle  a.     The  forces  acting  upon 
it  are  its  weight  and  the  pressure  ex- 
erted by  the  body  which  it  touches.    The  \^ 
former  is  mg  poundals  vertically  down-  \ 
ward,  and  the  latter  is  perpendicular  to  \     ^ 
the  plane  but   of  unknown  magnitude                            $ -J/^ 
(say  N  poundals).    The  force  N  is  called          •           <^^^ 
into  action  only  to  resist  a  tendency  of               ^<\  N 
the  body  to  pass  through  the  surface ;  JT^<^\  ,  \ 
it  cannot  cause  the  body  to  leave  the                    x  ma 
surface,  for  it  ceases  to  act  as  soon  as                    Fig.  135. 
the  bodies  separate.     The  motion  is  thus 

"  constrained"  to  follow  the  plane.  It  will  be  assumed  also  that  the 
motion  is  confined  to  a  vertical  plane,  which  is  taken  as  the  plane  of 
the  figure. 

Referring  to  rectangular  coordinates,  let  the  axis  of  x(OX,  Fig. 
135)  lie  in  the  plane,  the  positive  direction  being  downward  along 
the  plane.     The  equations  of  motion  are 

mx  =  mg  sin  a;     .....      (1) 

my  =  N  —  mg  cos  a.    .  .  .  .      (2) 

But  since  the  path  is  the  straight  line  y  =  o,  so  that  y  =  o,  equa- 
tion (2)  becomes 

N  —  mg  cos  a  =  o.  .         .         .     (3) 

Equation  (3)  serves  to  determine  N,  while  (1)  determines  the  motion. 


262 


THEORETICAL    MECHANICS. 


Comparing  equation  (i)  with  the  equation  of  motion  for  a  body 
falling  vertically,  it  is  seen  that  the  results  deduced  in  Art.  227  may 
be  applied  to  the  present  case  by  substituting  g  sin  a  for  g. 

The  three  cases  in  which  the  initial  velocity  is  zero,  down  the 
plane,  and  up  the  plane,  may  be  discussed  as  were  the  corresponding 
cases  in  Art.  227. 

306.  Motion  on  a  Rough  Plane  Under  Gravity. —  If  the  plane  is 
rough,  friction  must  be  included  among  the  forces  entering  the  equa- 
tions of  motion.  While  the  body  is  sliding,  the  force  of  friction  is 
equal  to  jjlN,  fi  being  the  coefficient  of  friction  (Art.  129)  ;  the  direc- 
tion of  this  force  is  opposite  to  the  sliding.  Taking  axes  as  in  Fig. 
136,  the  equations  of  motion  are 

mx  =  mg  sin  a  —  /jlN  ;  .  .  (1) 

my  =  N  —  mg  cos  a.  ...      (2) 

The  latter  equation  becomes 

N  —  mg  cos  a  =  o,     .  .         .         .     (3) 

because  of  the  condition  y  ==  o.      Equation  (3)  gives  the  value  of 
N  \  substituting  this  in  (1), 

x  ~  g{su\  a  —  fi  cos  a).         .  .  (4) 

The  integration  of  equation  (4)  gives  results  agreeing  with  those 
of  Art.  227,  with  the  substitution  o{g(s'm  a  —  /jl  cos  a)  for  g.     This 

equation  applies  only  to  the  case  in 
which  the  motion  is  down  the  plane.  If 
it  is  up  the  plane  the  direction  of  the 
force  /jlN  must  be  reversed.  If  x  is  still 
taken  as  positive  downward  along  the 
plane,  the  case  of  motion  up  the  plane 
gives  the  equation 

x  =  ^(sin  a  -j-  jjl  cos  a).    .     (5) 

It  should  be  noticed  that,  in  the  case 
of  motion  down  the  plane,  the  velocity 
will  increase  or  decrease,  according  as  the  second  member  of  (4)  is 
positive  or  negative.      Between  these  cases  is  that  in  which 

sin  a  —  jjl  cos  a  =  o,      or     /jl  ~  tan  a, 
which  reduces  equation  (4)  to  the  form 


x  =  o. 


PLANE    MOTION    OF    A    PARTICLE.  263 

This  is  the  case  of  equilibrium  ;  the  particle  will  remain  at  rest  or 
will  move  uniformly  down  the  plane. 

Uniform  motion  up  the  plane  is  impossible. 

Examples. 

1.  A  body  slides  down  a  smooth  plane  whose  inclination  to  the 
horizon  is  200.  How  far  does  it  move  during  the  first  2  sec.  after 
starting  from  rest?  Ans.   22  ft. 

2.  A  body  slides  down  a  smooth  plane  whose  inclination  to  the 
horizon  is  250.  What  is  the  velocity  when  it  reaches  a  position  12 
ft.  lower  than  its  position  of  rest?  Ans.   27.8  ft.-per-sec. 

3.  Solve  Ex.  2  if  the  inclination  of  the  plane  is  400,  the  data  being 
otherwise  unchanged. 

4.  A  body  starting  from  rest  slides  down  a  smooth  plane  whose 
inclination  to  the  horizon  is  a.  What  is  its  velocity  after  moving  a 
distance  whose  vertical  projection  is  k?  Show  that  the  velocity  ac- 
quired in  falling  to  a  given  level  is  independent  of  the  inclination  of 
the  plane. 

5.  A  body  is  projected  up  a  smooth  plane  inclined  300  to  the 
horizon  with  a  velocity  of  200  c.m.-per-sec.  What  is  its  velocity 
after  moving  40  c.  m.  ?  When  will  it  come  to  rest,  and  when  will  it 
return  to  the  point  from  which  it  is  projected? 

Ans.    28  c.m.-per-sec;  after  0.41  sec. ;  after  0.82  sec. 

6.  A  body  is  projected  with  a  velocity  V  up  a  smooth  plane 
whose  inclination  is  a.  When  will  it  come  to  rest,  when  will  it  return 
to  the  starting  point,  and  how  high  will  it  rise  ?  Prove  that  it  will 
rise  to  the  same  height  as  if  projected  vertically  upward  with  the 
same  speed. 

7.  Solve  Ex.  1  assuming  the  plane  rough,  the  angle  of  friction 
being  1  o°.  Ans.  11.4  ft. 

8.  A  body  is  projected  with  a  velocity  of  20  ft.  -per-sec.  down  a 
plane  whose  inclination  is  250,  the  coefficient  of  friction  being  0.4. 
Determine  the  position  and  velocity  after  2  sec. 

Ans.   43.9  ft.  from  starting  point ;  23.9  ft. -per-sec. 

9.  In  Ex.  8,  if  fi  =  o.  5,  when  will  the  body  come  to  rest  ? 

Ans.  After  20. 4  sec. 

10.  In  Ex.  8,  if  the  angle  of  friction  is  250,  determine  the  motion. 

11.  In  Ex.  8,  let  the  direction  of  the  initial  velocity  be  reversed, 
and  let  the  angle  of  friction  be  250.      Determine  the  motion. 

Ans.  The  body  will  come  to  rest  after  0.73  sec.  and  will  remain 
at  rest. 

1 2.  A  body  is  placed  at  rest  on  a  rough  plane,  and  the  inclina- 
tion of  the  plane  is  increased  until  sliding  begins.  If  the  coefficient 
of  friction  is  0.2,  what  is  the  angle  of  incipient  sliding? 


264  THEORETICAL    MECHANICS. 

13.  A  body  is  projected  up  an  inclined  plane.  What  condition 
must  be  satisfied  in  order  that  it  may  slide  down  after  coming  to  rest? 

14.  A  body  is  projected  up  a  plane  whose  inclination  to  the  hori- 
zon is  30°.  The  angle  of  friction  is  200.  If  the  initial  velocity  is  20 
ft. -per -sec. ,  (a)  when  will  the  body  come  to  rest,  (p)  when  will  it 
return  to  the  initial  position,  and  (r)  with  what  velocity  will  it  pass 
through  the  initial  position  ? 

Ans.  {a)  At  end  of  0.76  sec.  (b)  1.60  sec.  after  coming  to  rest. 
(<:)  9.52  ft.-per-sec. 

1 5.  A  body  is  projected  on  a  horizontal  plane  with  a  velocity  of 
50  ft.-per-sec.  It  comes  to  rest  after  6  sec.  What  is  the  degree  of 
roughness  of  the  plane? 

16.  A  stone  thrown  horizontally  on  a  sheet  of  ice  with  a  velocity 
of  1 20  ft. -per-sec.  comes  to  rest  after  sliding  1,400  ft.  Assuming 
the  coefficient  of  friction  to  be  independent  of  the  velocity,  determine 
its  value.  Ans.  0.16. 

17.  Write  the  equations  of  motion  of  a  body  sliding  on  a  smooth 
inclined  plane,  under  the  action  of  no  force  except  gravity  and  the 
pressure  of  the  plane,  assuming  that  the  motion  is  not  confined  to  a 
vertical  plane.     Show  that  the  path  is  a  parabola. 

18.  In  the  case  described  in  Ex.  17,  show  that  the  motion  is 
given  by  equations  like  those  deduced  in  Art.  295  for  a  projectile, 
with  the  substitution  of  g  sin  a  for  g,  a  being  the  inclination  of  the 
plane  to  the  horizon. 

307.  Motion  of  a  Bead  on  a  Smooth  Wire The  general  case 

in  which  a  particle  is  constrained  to  move  in  a  given  plane  curve  will 
now  be  considered.     To  fix  the  ideas,  let  the  particle  be  a  bead 

sliding  on  a  smooth  wire  bent  into  any 
1     I  plane  curve,  and  suppose  all  forces  act- 

/  ing  on  the  bead  to  be  known,  except 

qq  / X'     the  pressure  exerted  by  the  wire.     This 

/PST  _  pressure  is  a  "passive  resistance" 
yy  .'  ^\  (Art.  41);  it  comes  into  action  to  re- 
/a..  •  sist  any  tendency  of  the  bead  to  leave 

L LJ the  wire.     The  direction  of  the  pres- 

pIG  I37  sure  exerted  by  the  wire  upon  the  bead 

is  normal  to  the  path  of  the  particle. 
Let  N  represent  the  magnitude  of  this  unknown  normal  pressure, 
regarded  as  positive  if  directed  toward  the  concave  side  of  the  path. 
(See  Fig.  137.)  Let  Q  denote  the  resultant  of  all  forces  applied  to 
the  bead  except  the  normal  pressure  N. 

Two  independent  equations  of  motion  are  now  to  be  written,  ob- 


PLANE    MOTION   OF    A    PARTICLE.  265 

tained  by  resolving  resultant  force  and  resultant  acceleration  in  any 
two  directions.     Two  sets  of  equations  will  be  given. 

(1)  Resolution  along  fixed  rectangular  axes. — Choosing  any  pair 
of  rectangular  axes,  let  6  denote  the  angle  between  the  tangent  to  the 
path  and  the  axis  of  x.  Then  the  normal  pressure  N  makes  with 
the  ^r-axis  the  angle  900  +  6,  and  with  the  jj/-axis  the  angle  0  (Fig. 
137).     The  axial  components  of  A^  are  therefore 

—  N  sin  6     and     N  cos  6. 

But  if  s  means  the  length  of  the  path,  measured  from  some  fixed 
point,  sin  6  =  dy/ds,  cos  6  =  dx/ds  ;  therefore 

—  N(dy/ds)  =  .^-component  of  N; 

N{dxjds)  =jj/-component  of  N. 

Let  X\  Y'  denote  the  axial  components  of  the  known  force  Qt 
and  X,  Y  the  axial  components  of  the  resultant  of  all  forces  acting 
on  the  bead  (t.  e.,  the  resultant  of  Q  and  N).     Then 

X  ==  X'  —  N{dylds)  ; 

Y=  Y'  +  N(dx/ds). 

The  equations  of  motion  then  become  (Art.  287) 

m(d2x/dt2)  =  X'  —  Nidylds)  ;  .     (1) 

m(dy/df2)  =m  Y' .+  N(dx/ds).  .         .     (2) 

These  two  equations,  together  with  the  equation  of  the  path,  serve 
to  determine  the  motion. 

(2)  Resolution  along  tangent  and  normal. — Let  Qt  and  Qn  rep- 
resent the  tangential  and  normal  components  of  Q ;  then  if  forces 
and  acceleration  be  resolved  along  the  tangent  and  normal  to  the 
curve  at  each  instant  (Art.  289),  the  equations  of  motion  become 

m(dv/df)=Qt;  ....     (3) 

m{v2IR)  =N  +  Qn.         .         .         .     (4) 

In  equation  (4),  R  denotes  the  radius  of  curvature  of  the  path  at  the 
position  of  the  particle. 

Equations  (3)  and  (4)  are  just  equivalent  to  (1)  and  (2),  and 
either  set  may  be  used,  as  most  convenient. 

308.  Bead  Sliding  on  Smooth  Wire  Under  Gravity. —  Let  the 
resultant  applied  force  Q  be  constant  in  magnitude  and  direction.     If 


266  THEORETICAL    MECHANICS. 

the  ^r-axis  is  taken  in  the  direction  of  this  force,  X'  =  Q,  V  =  o. 
If  the  force  Q  is  the  weight  of  the  particle,  Q  =  mg,  and  the  equa- 
tions become 

m(d*xldt*)  =  mg  —  N{dylds)  ;  .         •     v5) 

m(d2y!dt2)  =  N(dx/ds).     ....     (6) 

An  important  result  may  be  deduced  from  these  equations,  irre- 
spective of  the  form  of  the  curve. 

Multiplying  (5)  by  dx  and  (6)  by  dy  and  adding,  there  results 

ldxd2x  dyd2y  ,\  , 

m dt  -f-  —  — -dt    =  mg  dx, 

\dtdf  dt  dt2     I  f. 

the  integration  of  which  gives 

fdx\2       1  (dy 


y       1  tdyy 
H —  — 1  =  gx  -f-  constant, 
2\dt)        Adt)        S 


or  v2  =  2gx  -\-  constant, 

if  v  is  the  speed.  Taking  the  integral  between  limits  xA  and  x2  for 
x,  and  limits  7\  and  v.,  for  v, 

V*  —  V*  r==  2^2  —  ^)  =  2^-//, 

if  ^5  is  the  height  through  which  the  body  falls  while  the  velocity 
changes  from  vx  to  vt . 

Comparing  this  result  with  that  of  Ex.  1,  Art.  227,  it  is  seen  that 
the  change  in  the  speed  of  the  particle  is  the  same  as  if  it  had  fallen 
freely  through  the  same  vertical  distance  under  the  action  of  gravity. 

The  further  consideration  of  this  problem  will  be  restricted  to  the 
case  in  which  the  path  of  the  particle  is  a  circle.  The  problem  then 
becomes  identical  with  that  of  the  motion  of  a  simple  pendulum. 

309.  Simple  Pendulum. —  A  particle  suspended  from  a  fixed 
point  by  means  of  a  weightless  rigid  rod  or  flexible  inextensible 
string,  and  acted  upon  by  no  forces  except  gravity  and  the  force 
exerted  by  the  rod  or  string,  is  called  a  simple  pendulum. 

Such  a  pedulum  exists  only  ideally.  An  actual  pendulum  con- 
sists of  a  body  suspended  by  a  bar  or  string  possessing  weight.  The 
motion  of  any  actual  pendulum,  if  frictional  resistances  be  neglected, 
is  the  same  as  that  of  some  simple  pendulum,  so  that  the  discussion 
of  the  ideal  simple  pendulum  is  of  practical  value  as  well  as  theoret- 
ical interest. 

Consider  then  the  case  of  a  particle  suspended  by  a  perfectly 


PLANE    MOTION    OF    A    PARTICLE.  267 

flexible  but  inextensible  string.  So  long  as  the  string  remains  tense 
the  path  must  be  a  circle  with  the  point  of  suspension  as  center  ;  and 
the  force  exerted  by  the  string  upon  the  particle  is  directed  toward 
the  center. 

Let  O  (Fig.  138)  be  the  point  of  suspension,  A  the  lowest  posi- 
tion occupied  by  the  particle  during  its 
motion,  and  B  the  position  at  any  in- 
stant. Represent  the  angle  A  OB  by 
0,  and  let  OA  =  /.  Also  let  m  de- 
note the  mass  of  the  particle,  and  N 
the  magnitude  of  the  force  exerted  by 
the  string,  its  direction  being  toward 
the  center.  The  only  other  force  act- 
ing upon  the  particle  is  its  weight,  a 
force  of  magnitude  mg,  directed  down- 
ward. Let  s  denote  the  length  of  the  arc  AB ;  then  s  =  16. 
(Both  6  and  s  will  be  negative  if  the  particle  passes  to  the  left  of  A.) 

Equations  (5)  and  (6)  of  Art.  308  are  now  applicable ;  but  it  will 
be  better  to  resolve  along  the  tangent  and  normal  to  the  path.  The 
equations  thus  take  the  forms  (3)  and  (4)  (Art.  307).  .  Resolving  in 
the  direction  of  the  tangent, 

m(d'2s/dt'2)  =  — mg  sin  6;  .         .     (7) 

and  resolving  in  the  direction  BO, 

m(v2/l)  =  N  —  mg  cos  6.     .         .         .     (8) 
Since  s  =  1 6,  (7)  may  be  written 

dt0/dtt=—  (g/l)  sin  0.         .  .  .      (9) 

The  integration  of  this  equation  determines  the  motion,  while  from 

(8)  N  may  be  determined  when  the  motion  is  known.  This  would 
constitute  the  complete  solution  of  the  problem.  The  complete  in- 
tegration of  (9)  will  not  be  shown,  since  it  involves  an  elliptic  integral. 
One  integration,  however,  may  readily  be  performed,  and  the  value 
of  N  determined.  If  the  range  of  the  motion  is  small  in  comparison 
with  the  radius  (i.  e.y  if  the  angle  6  is  small),  a  complete  approximate 
solution  may  be  made  which  is  practically  correct. 

( 1 )  Exact  partial  solution. — Mutiplying  both  members  of  equation 

(9)  by  dd  and  integrating,  there  results 

VJOidtf  =  (g/l)  cos  6  +  C. 


268  THEORETICAL    MECHANICS. 

To  determine  C,  let  a  be  the  value  of  6  when  the  particle  is  at  rest 
in  its  highest  position  ;  then  C  =  —  (g/0  cos  a,  and  the  equation 
becomes 

(dOldt)2  =  2(£-//)(cos  0  —  cos  a).       .  .     (io) 

The  velocity  in  any  position  is  therefore 

v  =  l(dO\dt)  =  V-2gl(cos  6  —  cos  a).  .      (i  i) 

The  value  of  N  may  now  be  determined  by  substituting  the  value 
of  v2  in  (8).     That  is, 

JV  =  m(v2/l)  -f-  mg  cos  6  =  w[2^(cos  0  —  cos  a)  -f  g  cos  0], 
or  AT  =  mg($  cos  6  —  2  cos  a).  .  .      (12) 

(2)  Approximate  complete  solution. — Suppose  the  angle  6  to  re- 
main very  small  throughout  the  motion  ;  then,  approximately, 

sin  6  =  s/l, 
and  equation  (7)  becomes 

d*s/dt*  =  -(gU)s (13) 

Multiplying  by  2{dsjdt)dt  and  integrating, 

(ds/dty  =  (g/l)(a*  -  i-2).     .  .  .      (14) 

Here  the  constant  of  integration  has  been  determined  on  the  as- 
sumption that  s  =  a  when  the  particle  is  at  rest  in  its  highest  position. 
Equation  (14)  may  be  written 

dtVJfl  =  ds/V 'a1  —  s2. 
Integrating,  and  determining  the   constant   by   the    condition    that 
s  =  o  when  /  =  o, 

tV g\l  =  sin-'O/tf)  —  sin-^o). 

Taking  sine  of  each  member  and  reducing,* 

s  =  asm(tV/g/iy  .         .         .     (15) 

This  result  is  identical  in  form  with  that  found  in  Art.  229  for  the 
motion  of  a  particle  in  a  straight  line  under  a  central  attractive  force 
varying  directly  as  the  distance.  The  motion  of  a  simple  pendulum 
when  the  range  is  small,  is  thus  a  harmonic  oscillation. 

*  It  will  be  noticed  that  the  conditions  assumed  in  determining  constants 
of  integration  leave  it  uncertain  whether  the  plus  or  the  minus  sign  should  be 
prefixed  to  the  second  member  of  (15).  If  it  be  further  assumed  that  s  is  plus 
for  small  positive  values  of  t  (i.  e.,  that  the  particle  is  moving  in  the  positive 
direction  through  A  when  t  =  o),  the  plus  sign  must  be  used.     See  Art.  229. 


PLANE    MOTION   OF    A   PARTICLE.  269 

Time  of  vibration  and  of  oscillation. —  If  C  and  C  (Fig.  139) 
are  the  extreme  positions  of  the  particle,  the  passage  from  C  to  C  or 
iirom  C  to  C  is  called  a  vibration,  while  the  passage  from 
C  to  C  and  back  to  C  is  called  an  oscillation.  The  par- 
ticle is  at  C  as  often  as  s  - -  a,  and  at  C  as  often  as  s  = 
— a.  While  s  changes  from  a  to  — a,  sin  (tvg/l) 
changes  from  +1  to  — 1,  and  tV g\l  changes  by  1800 
or  7r.     Hence  the  time  of  one  vibration  is 

T  =  Wlfg,      .        .        .     (16)  Qh&^c 

while  the  time  of  a  complete  oscillation  is  2T.  Fig.  139. 

Seconds  pendulum. —  A   seconds   pendulum  is  one 
that  vibrates  once  every  second.    Its  length  may  be  found  from  (16) 
if^-  is  known  ;  thus,  if  7"  =  1,  /  =  g/ir2. 

Determination  of  g. —  By  determining  the  time  of  vibration  of  a 
pendulum  of  known  length,  the  value  of  g  may  be  determined  from 
equation  (16). 

Examples. 

1.  What  is  the  length  of  the  seconds  pendulum  at  a  place  where 
the  value  of^*  is  32.2  ft.-per-sec.-per.-sec? 

2.  If  gx  and  g2  are  the  values  of  g  at  the  surface  of  the  earth  and 
at  an  elevation  of  h  ft.  above  the  surface  respectively,  what  is  the  rela- 
tion between  gx  and  g2  ? 

Ans.  If  Px  and  P2  are  the  values  of  the  earth's  attraction  upon 
the  body  in  the  two  positions,  the  law  of  gravitation  (Art.  176)  gives 
the  proportion  pjp%  —  rp  _l  k)2/R2. 

Or,  since  PJP2=gJg2, 

gJg>=(R  +  hyiR\ 

3.  If  h  is  small  compared  with  R,  show  that  the  result  of  Ex.  2 
reduces  approximately  to  the  form^2  —  gx(i  —  2/1 /R). 

4.  If  Tx  and  T.2  are  the  times  of  vibration  of  the  same  pendulum 
at  two  places  of  which  the  second  is  h  ft.  higher  than  the  first,  what 
is  the  relation  between  Tx  and  T2?  Show  that  this  relation  is  approx- 
imately expressed  by  the  equation  Tx  =  T.2(i  —k/R). 

5.  Show  how  the  last  result  may  be  used  in  determining  the  dif- 
ference of  elevation  between  two  points  in  the  same  neighborhood, 
the  same  pendulum  being  swung  in  the  two  places. 

6.  If  Nx  is  the  number  of  vibrations  of  a  pendulum  in  any  time  at 
a  certain  place,  and  N.2  the  number  of  vibrations  of  the  same  pendu- 
lum in  an  equal  time  at  a  place  h  ft  higher,  show  that,  approximately, 
kjR  =  (AT,  -  NMNV 


270  THEORETICAL    MECHA^JCS. 

7.  At  sea-level  a  pendulum  beats  seconds.  At  the  top  of  a  moun- 
tain it  beats  86, 360  times  in  24  hours.  What  is  the  height  of  the 
mountain  ? 

8.  Determine  roughly  a  value  of  g  by  swinging  a  weight  sus- 
pended by  a  string. 

9.  A  body  whose  mass  is  1  lb.  is  suspended  from  a  fixed  point 
by  a  string  12  ft.  long.  The  string  is  swung  to  a  position  6o°  from 
the  vertical  and  the  body  released.  Determine  the  velocity  when 
the  body  is  in  its  lowest  position ;  also  when  2  ft.  above  its  lowest 
position. 

10.  In  Ex.  9,  determine  the  tension  in  the  string  in  each  of  the 
positions  specified  ;  also  in  the  initial  position. 

Ans.  Tension  in  lowest  position  is  twice  the  weight  of  the  body  ; 
in  highest  position,  half  the  weight  of  the  body. 

1 1.  Two  bodies  A  and  B  are  connected  by  a  flexible,  inextensible 
string  which  passes  over  a  smooth  pulley  C.  The  body  A  is  verti- 
cally below  C  and  rests  on  a  horizontal  plane.  The  body  B  is  held 
on  a  level  with  C,  the  string  being  tight,  and  is  then  allowed  to  fall 
under  the  action  of  gravity.  How  far  will  it  fall  before  A  leaves  the 
floor? 

Ans.  Let  mf  m'  be  the  masses  of  A  and  B  respectively,  and  let 
6  =  A  CB.  Then  at  the  instant  when  A  is  lifted,  cos  6  =  m/$m'. 
If  m  >  3#z',  the  body  A  will  not  be  lifted. 

310.  Uniform  Circular  Motion. —  If  a  particle  describes  a  circle 
of  radius  r  with  uniform  speed  v,  its  acceleration  is  at  every  instant 

directed  toward  the  center  of  the  circle, 
and  is  equal  to  v2/r.  Therefore,  from 
the  general  equation  of  motion,  the  re- 
sultant of  all  forces  acting  upon  the  par- 
ticle is  a  force  mv'2/r  directed  toward 
the  center. 

Let  A    (Fig.  140)   be  the  particle, 
and  suppose  it  to  be  attached  to  an  in- 
extensible  string  A  O,  of  which  the  end 
O  is  fastened  to  a   fixed   peg.      If  the 
Fig.  140.  particle  be  projected  in  a  direction  at 

right  angles  to  A  O  with  a  velocity  vt 
and  if  no  force  acts  upon  it  except  that  due  to  the  string,  the  equa- 
tions of  motion  obtained  by  resolving  along  the  tangent  and  normal 
to  the  path  (Art.  289)  are 

m(dv\df)  =  o,     m(y*/r)  =  T; 


PLANE    MOTION    OF    A    PARTICLE.  27 1 

T  being  the  force  exerted  upon  the  particle  by  the  string.  The  first 
equation  shows  that  the  speed  v  remains  constant ;  the  second  gives 
the  value  of  T.  The  string  thus  exerts  upon  the  particle  a  force 
whose  direction  is  always  toward  the  point  O  and  whose  magnitude 
is  mv%\r.  By  the  law  of  action  and  reaction,  the  particle  exerts  upon 
the  string  a  force  mv^fr  in  the  direction  OA.  The  string  sustains  a 
tension  whose  value  is  uniform  throughout  its  length  ;  i.  e.,  if  it  be 
conceived  to  be  divided  by  a  section  at  any  point,  the  two  portions 
exert  upon  each  other  forces  which  are  equal  and  opposite  (Art.  43), 
each  being  equal  to  mv2jr.  The  peg  is  pulled  by  the  string  in  the 
direction  OA  with  a  force*  equal  to  mva/r. 

The  value  of  the  resultant  force  acting  upon  the  particle  may  be 
expressed  in  terms  of  its  angular  velocity  (Art.  281)  about  the  center 
of  the  circle.     If  the  angular  velocity  \s  co,  v  —  rco,  and 

mva/r  =  mrco'\ 

Examples. 

1.  A  body  of  2  lbs.  mass  is  swung  in  a  circle  of  2  ft.  radius  by  a 
string  which  is  capable  of  sustaining  2  lbs.  against  gravity.  At  what 
rate  may  it  move  without  breaking  the  string  ? 

Ans.  About  8  ft.-per-sec. 

2.  A  locomotive  of  250,000  lbs.  mass  describes  a  curve  of  2,000 
ft.  radius  at  the  rate  of  30  miles-per-hour.  What  is  the  resultant 
force  acting  upon  it  ?  What  horizontal  pressure  does  it  exert  upon 
the  rail?     Ans.   Resultant  force  =  7,520  lbs.  toward  the  center. 

3.  A  locomotive  of  mass  m  describes  a  curve  of  radius  r  with 
speed  v.  (a)  Determine  the  resultant  force  acting  on  the  locomo- 
tive, (fi)  Determine  the  magnitude  and  direction  of  the  resultant 
pressure  between  the  track  and  the  locomotive,  (r)  What  should  be 
the  difference  in  elevation  of  the  rails  in  order  that  there  shall  be  no 
tendency  of  the  locomotive  to  slide  laterally? 

Ans.  (a)  The  resultant  force  acting  upon  the  locomotive  is  a 
force  mv2!r  directed  toward  the  center  of  the  curve.  This  resultant 
is  made  up  of  two  components, —  a  downward  force  mg  (the  weight 
of  the  body)  and  the  supporting  force  exerted  by  the  track.  (6) 
This  supporting  force  assumes  such  magnitude  and  direction  that, 
when  combined  with  the  weight,  it  gives  the  resultant  mv*/r  toward 
the  center.     Hence  its  magnitude  is 

*  This  force  acting  upon  the  peg  is  sometimes  called  the  "centrifugal" 
force.  It  seems  better,  however,  to  reserve  this  term  for  another  use,  in  con- 
nection with  problems  in  relative  motion. 


272  THEORETICAL    MECHANICS. 


V(mv*/ry  +  {pigf  =  mVfy*jr)%  +  g'z; 
and  its  inclination  to  the  vertical  is  tan-1  (v2jrg).  (c)  The  pressure 
between  the  rails  and  the  locomotive  should  be  normal  to  the  surface 
determined  by  the  tops  of  the  rails ;  hence  this  surface  should  make 
with  the  horizontal  the  angle  whose  tangent  is  v2/rg.  If  the  hori- 
zontal distance  between  centers  of  rails  is  a,  the  outer  rail  should  be 
higher  than  the  inner  by  v2a/gr. 

4.  A  horizontal  platform  rotates  about  a  vertical  axis  at  the  uni- 
form rate  of  6  revolutions  per  min.  A  body  of  20  lbs.  mass  rests 
upon  the  platform  at  a  point  16  ft.  from  the  axis  of  rotation,  (a) 
What  is  the  resultant  force  acting  upon  the  body  ?  (b)  Of  what  ac- 
tual forces  is  this  resultant  composed?  (c)  How  great  must  be  the 
coefficient  of  friction  to  prevent  the  body  from  sliding  ? 

Ans.  (a)  126.3  poundals  toward  the  axis  of  rotation,  (b)  The 
actual  forces  are  the  weight  (20^  poundals)  and  the  pressure  of  the 
supporting  platform.  The  normal  component  of  this  pressure  is  20^ 
poundals  upward,  the  tangential  component  is  126.3  poundals  toward 
the  axis  of  rotation.  (c)  The  coefficient  of  friction  must  be  at  least 
126.3/20^*  =  o.  196. 

5.  In  Ex.  4,  if  the  coefficient  of  friction  is  o.  5,  how  far  from  the 
axis  may  the  body  be  placed  without  sliding?  Ans.  40.8  ft. 

6.  In  Ex.  4,  if  the  body  rests  upon  a  smooth  surface  which  is 
fixed  to  the  platform,  what  must  be  the  slope  of  the  surface? 

7.  In  Ex.  4,  if  the  body  is  suspended  by  a  string  from  a  support 
fixed  to  the  platform,  what  direction  will  the  string  assume,  and  what 
tension  will  it  sustain? 

Ans.  The  lower  end  of  the  string  will  swing  directly  away  from 
the  axis  of  rotation  until  the  string  is  inclined  n°  6'  to  the  vertical. 
The  tension  will  be  656  poundals. 

8.  A  platform  rotates  about  a  vertical  axis  with  uniform  angular 
velocity  co.  A  body  distant  x  from  the  axis  is  placed  on  a  smooth 
surface  whose  inclination  is  such  that  the  body  rests  without  sliding. 
Show  that  the  surface  is  inclined  to  the  horizontal  at  an  angle  whose 
tangent  is  xco2!g.     The  surface  slopes  directly  toward  the  axis. 

9.  Using  the  result  of  Ex.  8,  show  that  a  smooth  surface  upon 
which  a  body  would  rest  wherever  placed  must  be  a  surface  gener- 
ated by  a  parabola  which  rotates  with  the  platform  and  whose  prin- 
cipal diameter  lies  in  the  axis  of  rotation. 

If  x,  y  are  the  coordinates  of  a  section  of  the  surface  by  a  plane 
containing  the  axis  of  rotation,  the  origin  being  at  the  point  in  which 
the  axis  pierces  the  surface,  the  result  of  Ex.  8  gives 

dy/dx  =  (co2/g)x. 
Integrating,  y  =  (co2/2g)x2. 

10.  Two  bodies,  each  of  5  lbs.  mass,  are  connected  by  an  elastic 
string  which  passes  through  a  straight  tube  3  ft.  long  and  by  its 


PLANE    MOTION    OF    A    PARTICLE.  273 

tension  holds  the  bodies  against  the  ends  of  the  tube.  The  system 
rotates  uniformly  about  a  vertical  axis  perpendicular  to  the  axis  of 
the  tube  at  its  middle  point.  The  natural  length  of  the  string  is 
such  that  a  pull  equal  to  the  weight  of  2  lbs.  is  required  to  stretch  it 
to  the  length  of  the  tube.  What  is  the  pressure  of  each  body  against 
the  tube  when  the  system  is  making  10  revolutions  per  minute? 
(Take  g  =  32.2  ft.-per-sec.-per-sec.) 

Ans.  56.2  poundals  or  1.75  pounds-force. 

11.  Let  the  system  described  in  Ex.  10  rotate  about  a  vertical 
axis  6  ins.  from  the  middle  point  of  the  tube.  Determine  the  pres- 
sure of  each  body  against  the  tube. 

Ans.  The  body  nearer  the  axis  presses  against  the  tube  with  a 
force  of  58.9  poundals  ;  the  other  with  a  force  of  53.4  poundals. 

12.  The  system  and  axis  of  rotation  being  as  described  in  Ex.  10, 
suppose  the  angular  velocity  gradually  to  increase.  When  will  the 
bodies  cease  to  press  against  the  tube? 

Ans.  When  the  angular  velocity  reaches  2.93  rad. -per-sec. 

13.  If  the  axis  of  rotation  is  located  as  in  Ex.  n,  and  the  angular 
velocity  gradually  increases,  which  body  will  leave  the  tube  first? 
When  this  occurs,  what  will  be  the  pressure  of  the  other  body  on  the 
tube? 

Ans.  When  the  angular  velocity  reaches  2.54  rad. -per-sec.  the 
body  2  ft.  from  the  axis  of  rotation  will  cease  to  press  against  the 
tube.  The  pressure  of  the  other  against  the  tube  will  then  be  equal 
to  half  the  tension  in  the  string  or  32.2  poundals. 

14.  A  particle  suspended  from  a  fixed  point  by  an  inextensible 
string  and  acted  upon  by  no  force  except  its  weight  and  the  pull  of 
the  string  describes  a  horizontal  circle.  If  a  is  the  length  of  the 
string,  6  the  angle  it  makes  with  the  vertical,  a>  the  angular  velocity, 
T  the  tension  in  the  string,  and  m  the  mass  of  the  particle,  show  that 
cos  6  =  g/aa?,  T  =  maw2. 

15.  In  Ex.  14  let  the  mass  of  the  particle  be  1  kilogr.  and  the 
length  of  the  string  1  met.  If  the  number  of  revolutions  per  minute 
is  40,  determine  6  and  T.      (Assume  £"  =  981  C.  G.  S.  units.) 

Ans.  6  =  560  (nearly)  ;  T  =  1.755  X  io6  dynes  ==  1.788  kilo- 
grams-weight. 

311.  Effect  of  the  Earth's  Rotation  Upon  Apparent  Weights 
of  Bodies. —  If  by  the  weight  of  a  body  is  meant  the  gravitational 
pull  of  the  earth  upon  it,  the  apparent  weight  differs  from  the  true 
weight  because  of  the  earth's  rotation. 

Let  Fig.  141  represent  a  meridian  of  the  earth,  NS  being  the 

polar  diameter  and  B  a  point  in  the  equator.     The  form  of  the 

meridian  is  known  to  be  nearly  elliptical,  the  length  of  the  polar 

radius  being  6.356  X  io8  cm.   and  that  of  the  equatorial  radius 

18 


274 


THEORETICAL    MECHANICS. 


6.378  X  io8  cm.,  very  nearly.  In  Fig.  141  the  ellipticity  is  greatly 
exaggerated.  If  the  earth  were  spherical  and  of  uniform  density,  or 
of  density  varying  only  with  the  distance  from  the  center,  the  attrac- 
tion upon  a  body  at  the  surface  would  be  directed  toward  the  center 
(Art.  183).  In  fact  the  direction  differs  from  this.  In  Fig.  141,  let 
AH'  be  the  direction  of  the  earth's  attraction  upon  a  body  at  A, 
and  let  P'  denote  the  magnitude  of  the  attraction  upon  a  body  of 
mass  m. 

To  fix  the  ideas,  suppose  the  body  to  be  suspended  by  a  string 
A  C  attached  to  a  spring  balance,  and  let  P  be  the  supporting  force 
as  measured  by  the  balance.     If  the  body  were  in  equilibrium,  P 


Fig.  141. 

would  be  equal  and  opposite  to  P .  But  since  the  body  moves  with 
the  earth  in  its  daily  rotation,  the  string  assumes  such  a  direction  and 
the  supporting  tension  such  a  magnitude  that  the  resultant  of  P  and 
P'  is  just  sufficient  to  maintain  the  actual  motion  of  the  body. 

Let  r  —  radius  of  circle  described  by  body,  co  =  angular  velocity 
of  earth's  rotation ;  then  the  resultant  force  acting  upon  the  body 
has  the  direction  AD  and  the  magnitude  Q  =  mrco2.  The  value  of 
Q  may  thus  be  computed,  since  w  is  a  known  constant  and  r  is 
known  for  any  latitude. 

The  relation  between  P,  P'  and  Q  is  represented  by  the  vector 
triangle  EFGf  in  which  EF is  parallel  to  AH',  FG  to  AC,  and  EG 
to  AD.  The  value  of  P  is  determined  by  experiment ;  Q  may  be 
computed  as  above  ;  and  P'  may  then  be  determined  by  solving  the 
triangle  EFG. 

The  force  P  is  equal  and  opposite  to  the  '  ■  apparent ' '  weight  of 


PLANE    MOTION    OF    A    PARTICLE.  275 

the  body.  Its  direction  is  normal  to  the  earth's  surface  at  A  {i.  e.y 
to  a  level  surface,  such  as  the  surface  of  still  water).  The  angle  <f> 
between  A  C  and  the  plane  of  the  equator  is  the  latitude  of  the  place. 
It  will  be  shown  presently  (examples  3  and  4  below)  that  Q  is  very 
small  in  comparison  with  P,  so  that  the  angle  EFG  is  also  very 
small ;  assuming  this  to  be  true,  the  value  of  the  angle  and  of  the 
difference  between  P  and  P'  may  be  determined  to  a  close  approx- 
imation in  the  following  simple  form  : 

angle  EFG  =  /3  =  (Q  sin  <j>)/P;  .         .     (1) 

P'  —  P  =  Q  cos  cf>.        .         .         .     (2) 
Here  ft  is  expressed  in  radians. 

Examples. 

1.  The  time  of  one  revolution  of  the  earth  is  86,164  mean  solar 
sec.     Determine  the  value  of  od,  the  angular  velocity,  in  rad.  -per-sec. 

Ans.  log  (o  =  5.8628  —  10 ;  co  =  0.00007292. 

2.  Determine  the  linear  velocity  of  a  body  at  sea  level  at  the 
equator,  using  the  value  of  the  equatorial  radius  given  above. 

Ans.  4.65  X  io4  cm. -per-sec. 

3.  Determine  the  difference  between  the  true  and  the  apparent 
weight  of  a  body  of  m  gr.  at  the  equator. 

Ans.  For  a  body  at  the  equator  P,  P'  and  Q  are  all  parallel,  and 
P'  —  P=  Q.  Also,  Q  =  mv2/r,  in  which  v  =  4.65  X  io4  (Ex. 
2),  r  =  6.378  X  io8.     Therefore  Q  —  3.391^  dynes. 

4.  If  the  value  of^  at  sea-level  at  the  equator  is  978.1  c.  m. -per- 
sec. -per-sec. ,  what  would  be  its  value  if  not  modified  by  the  earth's 
rotation  ? 

The  value  of  g  as  experimentally  determined  is  the  ■ '  apparent ' ' 
acceleration, — i.  e.,  the  acceleration  on  the  assumption  that  the  earth 
is  at  rest.     If  P  is  the  apparent  weight  of  a  body  of  mass  m, 

g  =  Pjm. 
The  acceleration  which  would  be  due  to  the  true  attraction  is  Pf/m. 
At  the  equator, 

P'/m  =  Pjm  +  Qjm  =  978. 1  +  3.4  ==  981.5. 

5.  Show  that  if  the  earth  should  rotate  seventeen  times  as  rapidly 
as  at  present,  its  form  and  density  being  unchanged,  the  apparent 
weights  of  bodies  at  the  equator  would  be  reduced  nearly  to  zero. 

6.  In  latitude  400  the  radius  of  a  parallel  of  latitude  is  very  nearly 
4.894  X  io8  cm.  The  value  of^-at  sea-level  is  about  980.2  C.  G.  S. 
units.  Determine  the  influence  of  the  earth's  rotation  on  the  mag- 
nitude and  direction  of  g. 


276  THEORETICAL    MECHANICS. 

Using  C.  G.  S.  units,  the  value  of  Q  is  mrooi1,  in  which  r  =  4.894 
X  io8,  ft)  =  7.292  X  io-5  (Ex.  1)  ;  that  is,  Q  =  2.602m  dynes. 
From  equation  (2), 

P'  —  P  =  Q  cos  <j>  --=  2.602m  cos  400  =  1.993W. 
From  the  given  data,  P  =  980. 2m,  hence  P'  =  982. 2m  ;  and  equa- 
tion (1)  gives 

/3=(Q  sin  <j>)/P'  =  (2.602  sin  4o°)/982.2 
==  0.001703  radians  =  5'  51". 

The  rotation  of  the  earth  thus  changes  the  magnitude  of  g  by  1.993 
C.  G.  S.  units,  and  its  direction  by  5'  51". 

7.  If  the  length  of  a  degree  of  longitude  is  78,849  meters  in  lati- 
tude 450,  and  the  value  of  g  980. 6  C.  G.  S.  units,  compute  the  effect 
of  the  earth's  rotation  on  the  value  of  g. 

8.  If  the  diminution  of  the  apparent  weight  of  a  given  body  in 
latitude  cf>  is  w,  and  if  the  value  of  w  at  the  equator  is  w\  show  that 
w  =  w'  cos2  <j>  (very  nearly  J. 


CHAPTER    XVL 

MOMENTUM    AND    IMPULSE. 

§  i.   Rectilinear  Motion. 

312.  Momentum. —  The  fnomentum  of  a  particle  is  a  quantity 
proportional  directly  to  its  mass  and  to  its  velocity. 

Momentum  is  sometimes  called  mass-velocity.  It  is  often  re- 
garded as  a  measure  of  the  "  quantity  of  motion  "  of  the  particle. 

The  numerical  value  of  the  momentum  of  a  particle  of  given 
mass  moving  with  given  velocity  depends  upon  the  unit  in  terms  of 
which  it  is  expressed. 

Unit  of  momentum. —  For  most  purposes  it  is  convenient  to 
choose  as  the  unit  the  momentum  possessed  by  a  body  of  unit  mass 
having  the  unit  velocity.  The  unit  thus  defined  depends  upon  the 
units  of  mass,  length  and  time.  If  these  are  the  pound,  the  foot 
and  the  second,  respectively,  the  unit  of  momentum  is  that  of  a  body 
of  one  pound  mass  having  a  velocity  of  one  foot-per-second. 

Other  units  would  be  the  momentum  of  a  particle  of  one  gram 
mass  having  a  velocity  of  one  centimeter-per-second  ;  and  that  of 
a  kilogram  mass  having  a  velocity  of  one  meter-per-second. 

In  the  discussion  which  immediately  follows,  it  is  to  be  under- 
stood that  the  motion  is  restricted  to  a  straight  line. 

313.  Increment  of  Momentum. —  If  the  velocity  of  a  particle 

varies,  so  also  does  the  momentum.      If  the  mass  is  tn,  and  if  the 

velocity  has  values  vx  and  vt  at  the  beginning  and  end  of  a  given 

interval  respectively,  the  increment  of  the  momentum  during  the 

interval  is  ,  -. 

mv2  —  mvx  =  m{v2  —  vj. 

314.  Acceleration  of  Momentum. —  The  rate  of  increase  of  the 
momentum  (or  increment  of  momentum  per  unit  time)  is  called  the 
acceleration  of  momentum. 

If  the  velocity  varies  at  a  uniform  rate,  receiving  an  increment  Av 
in  an  interval  of  time  At,  the  acceleration  of  momentum  is  equal  to 
m(Av/Af).  If  the  velocity  changes  at  a  variable  rate,  the  instanta- 
neous value  of  the  acceleration  of  momentum  is  m(dv/dt).  In  either 
case  its  value  is  mp,  p  being  the  acceleration.  For  this  reason  it  is 
often  called  mass-acceleration. 


278  THEORETICAL    MECHANICS. 

The  most  convenient  unit  of  mass-acceleration  is  that  of  a  unit 
mass  having  unit  acceleration.  With  the  usual  British  units  of  mass, 
length  and  time,  this  unit  would  be  the  mass-acceleration  of  a  pound- 
mass  having  an  acceleration  of  one  foot-per-second-per-second.  If 
the  gram  and  centimeter  are  taken  as  units  of  mass  and  length  re- 
spectively, the  unit  mass-acceleration  is  that  possessed  by  a  particle 
of  one  gram  mass  having  an  acceleration  of  one  centimeter-per-sec- 
ond-per-second. 

315.  Impulse  of  a  Constant  Force. —  The  impulse  of  a  force 
which  remains  constant  in  direction  and  magnitude  is  a  quantity  pro- 
portional directly  to  the  force  and  to  the  time  during  which  it  acts. 
The  numerical  value  of  the  impulse  of  a  given  force  acting  for  a  given 
time  depends  upon  the  unit  in  terms  of  which  it  is  expressed. 

Unit  impulse. —  Let  the  unit  impulse  be  that  of  a  unit  force  act- 
ing for  a  unit  time.  Then  the  value  of  any  impulse  is  equal  to  the 
product  of  the  force  into  the  time. 

This  unit  depends  upon  the  units  of  force  and  of  time.  In  the 
F.  P.  S.  system  the  unit  impulse  is  that  of  a  poundal  force  acting  for 
a  second.  In  the  C.  G.  S.  system  it  is  the  impulse  of  a  dyne  acting 
for  a  second.  These  may  be  designated  as  a  poundal-second  and  a 
dyne-second  respectively.  If  the  engineers'  system  (Art.  218)  is  em- 
ployed, the  unit  impulse  is  the  pound-second. 

Impulse  is  a  localized  vector  quantity  (Art.  26),  its  direction  and 
line  of  action  coinciding  with  those  of  the  force.  Restricting  the 
discussion  to  forces  acting  upon  a  particle  moving  in  a  straight  line 
which  is  the  line  cf  action  of  all  the  forces,  the  only  directions  to  be 
distinguished  are  the  two  opposite  directions  along  the  path  ;  these 
may  be  distinguished  by  signs  +  and  — . 

316.  Impulse  of  a  Variable  Force. —  If  the  magnitude  of  a  force 
varies  continuously  throughout  a  certain  interval,  its  impulse  must  be 
computed  by  integration. 

Let  the  value  of  the  force  at  any  instant  t  be  denoted  by  P,  and 
let  it  be  required  to  determine  the  value  of  the  impulse  during  the 
interval  from  /  =  t'  to  t  =  t" ,  Let  this  interval  be  divided  into 
small  intervals  each  equal  to  At  ;  and  let  Px ,  P2 ,  .  .  .  denote 
the  values  of  P  at  the  beginnings  of  the  successive  small  intervals. 
Represent  the  required  impulse  by  P' ;  then,  approximately, 

P'  =  PlAt  +  Ps*t+ (0 


MOMENTUM    AND    IMPULSE.  279 

The  approximation  is  closer  the  smaller  At  is  taken.  If  At  is  made 
to  approach  zero,  the  number  of  terms  in  the  series  (i)  increases  in- 
definitely while  the  value  of  every  term  approaches  zero.  The  sum 
in  general  approaches  a  definite  value  which  is  the  exact  value  of  the 
impulse  P' .     That  is, 

P'  =  limit  [Px  At  +  P2  At  +     .     .     .     ]  =  C" Pdt.    .    (2) 

When  P  is  constant  this  reduces  to 

P  =  Pit"  -  /')> 

agreeing  with  Art.  315. 

317.  Relation  Between  Impulse  and  Momentum. —  If  m  is  the 
mass  of  a  particle  acted  upon  by  a  single  force  P  directed  along  the 
line  of  motion,  the  general  equation  of  motion  is 

P  =  m(dv/dt). 

Multiplying  through  by  dt  and  integrating  between  limits  /'  and  t'\ 


x 


Pdt  =  m(v"  —  v[),         .  .  •     (3) 


v'  and  v"  being  the  values  of  the  velocity  at  the  instants  t'  and  t"  re- 
spectively. 

The  first  member  of  equation  (3)  is  the  value  of  the  impulse  of 
the  force  P  during  the  interval  from  /'  to  t" .  In  the  second  member, 
mv'  is  the  value  of  the  momentum  at  the  instant  /',  and  mv"  its  value 
at  the  instant  t";  m(z>"  —  v')  is  the  increment  of  the  momentum  for 
the  interval  from  t'  to  t" .  Equation  (3)  therefore  expresses  the  prop- 
osition that 

The  impulse  of  a  force  acting  alone  upon  any  particle  during  a 
given  interval  is  equal  to  the  change  of  momentum  of  the  particle 
during  that  interval. 

If  several  forces  act  upon  the  particle,  its  change  of  momentum 
during  any  interval  is  equal  to  the  impulse  of  the  resultant  force ; 
and  this  is  evidently  equal  to  the  algebraic  sum  of  the  impulses  of  the 
several  forces. 

Examples. 

1.  A  body  whose  mass  is  m  pounds  falls  vertically  under  the 
action  of  gravity,  (a)  Compute  the  impulse  of  the  force  during  t 
sec.     (J?)  What  increment  of  momentum  is  received  during  t  sec? 

Ans.   (a)  mgt  poundal-seconds.     (J?)  mgt  momentum-units. 


28o  THEORETICAL    MECHANICS. 

2.  In  Ex.  i,  if  the  body  has  initially  an  upward  velocity  vx ,  what 
is  its  final  velocity?  Ans.  gt  —  v^  ft.-per-sec.  downward. 

3.  A  body  of  12  lbs.  mass  is  projected  downward  with  a  velocity 
of  20  ft.  -per-sec.  Write  the  equation  of  impulse  and  momentum  for 
an  interval  of  3  sec,  and  thus  determine  the  velocity  at  the  end  of 
the  interval. 

4.  A  particle  of  mass  m  is  acted  upon  by  a  force  P  =  kt,  t  being 
the  time  reckoned  from  a  definite  instant.  Compute  the  impulse 
during  the  interval  from  t  =  o  to  /  =  t' .  Write  the  equation  of  im- 
pulse and  momentum,  assuming  the  velocity  to  be  zero  when  /  =  o. 
Explain  the  meaning  of  k. 

5.  In  Ex.  4,  let  m  =  500  gr.,  and  let  P=  10,000  dynes  when 
/  =  1  sec.  Write  the  equation  of  impulse  and  momentum  and  deter- 
mine the  velocity  when  t  =  8  sec. 

Ans.   v"  =  640  cm. -per-sec. 

318.  Dimensions  of  Impulse  and  Momentum. —  If  units  of  force, 
mass,  length  and  time  are  all  chosen  arbitrarily,  the  unit  impulse  is 
of  dimensions  FT,  and  the  unit  momentum  of  dimensions  ML/T. 
But  with  such  an  arbitrary  choice  of  units  the  equation  of  impulse 
and  momentum  must  contain  a  constant,  taking  the  form 

(impulse)  =  (constant)  X  (change  of  momentum), 

the  value  of  the  constant  being  determined  in  accordance  with  the 
particular  fundamental  units  chosen. 

If,  as  in  Art.  317,  a  kinetic  system  of  units  is  employed,  the  unit 
force  has  dimensions*  ML/T2,  and  the  unit  impulse  is  therefore  of 
dimensions  ML/T.     The  equation 

impulse  =  change  of  momentum 

thus  becomes  homogeneous,  both  members  being  of  the  same  dimen- 
sions in  terms  of  the  fundamental  units  M,  L  and  T. 

319.  Sudden  Impulse. —  The  impulse  of  a  force  which  acts  for  a 
very  short  time  is  called  a  sudden  impulse.  If  the  time  of  action  is 
infinitesimal,  the  impulse  is  instantaneous. 

The  value  of  a  sudden  impulse  will  be  small  unless  the  force  is 
very  great.  For  PAt  is  the  impulse  of  a  constant  force  P  acting  for 
a  time  At,  and  this  product  will  be  very  small  if  At  is  very  small, 
unless  P  is  very  great. 

An  instantaneous  impulse  cannot  have  a  finite  value  unless  the 

*  See  Art.  219. 


MOMENTUM    AND    IMPULSE.  28 1 

force  is  infinite ;  impulses  which  are  strictly  instantaneous  therefore 
do  not  come  within  our  experience.  There  are,  however,  cases  in 
which  a  very  great  force  acts  for  a  very  short  time,  and  in  which  the 
variation  of  the  force  within  the  interval  considered  cannot  be  made 
the  subject  of  observation.  Such  impulses  may  often  be  regarded  as 
instantaneous  without  important  error.  Thus,  when  two  bodies  come 
into  collision,  the  force  which  each  exerts  upon  the  other  cannot  usu- 
ally be  determined,  since  its  time  of  action  is  very  short ;  but  the 
value  of  the  impulse  may  sometimes  be  found  by  observing  its  effect 
upon  the  motion. 

320.  Effect  of  a  Sudden  Impulse. —  The  effect  of  a  sudden  im- 
pulse is  to  produce  a  sudden  change  in  the  momentum  of  the  body 
acted  upon.  If  this  change  of  momentum  can  be  observed,  the  value 
of  the  impulse  can  be  determined,  although  the  force  and  the  time 
both  remain  unknown. 

321.  Ordinary  Forces  Neglected. —  In  considering  the  effect  of 
a  sudden  impulse,  all  forces  of  ordinary  magnitude  may  usually  be 
neglected  without  important  error.  Consider  the  case  of  a  ball  struck 
by  a  bat.  During  the  impulse  the  ball  is  acted  upon  by  the  force  of 
gravity  (a  downward  force  equal  to  the  weight  of  the  ball).  But 
during  the  short  time  of  action  of  the  force  due  to  the  blow,  the 
change  of  momentum  due  to  the  force  of  gravity  is  exceedingly  small 
in  comparison  with  that  due  to  the  blow,  and  may  be  neglected  with- 
out appreciable  error. 

If  an  impulse  is  strictly  instantaneous,  any  finite  force  whatever 
may  be  neglected  in  comparison  with  the  force  of  the  impulse,  since 
the  momentum  due  to  a  finite  force  during  an  infinitesimal  time  is 
infinitesimal. 

Let  P  be  the  force  of  an  instantaneous  impulse  which  produces  a 
finite  change  of  momentum,  and  let  Q  be  a  finite  force  acting  on  the 
particle  at  the  time  of  the  blow.     Then  the  equation  of  impulse  and 

momentum  is 

rt"  rt" 

Pdt  +  I     Qdt=  miro"  —  v'\ 
J  t<  J  t, 

If  the  time  of  action  of  the  blow,  /"  —  t\  approaches  zero,  the 
second  term  of  the  first  member  of  this  equation  approaches  zero. 
Hence  Q  may  be  neglected  in  estimating  the  change  of  momentum 
during  the  blow. 


282  theoretical  mechanics. 

Examples. 

i.  A  ball  weighing  5^  oz.,  moving  horizontally  at  the  rate  of 
100  ft. -per-sec. ,  is  struck  by  a  bat.  Immediately  after  the  blow  the 
ball  moves  at  the  rate  of  1 50  ft.  -per-sec.  in  the  direction  opposite  to 
that  of  its  original  motion.     What  is  the  value  of  the  impulse? 

2.  In  Ex.  1,  if  the  time  occupied  by  the  blow  is  o.  1  sec,  what  is 
the  average  value  of  the  force  (z.  e. ,  what  constant  force  would  pro- 
duce the  same  change  of  momentum  in  the  same  time)  ? 

Ans.  820.3  poundals. 

3.  What  would  be  the  velocity  of  the  ball  after  the  blow,  if  initially 
at  rest  and  acted  upon  by  the  same  impulse  as  in  Ex.  1  ? 

4.  If  the  same  blow  were  applied  to  a  body  of  1  lb.  mass,  what 
would  be  its  effect? 

322.  Impact  or  Collision  of  Bodies. —  When  two  bodies  come 
into  collision,  each  exerts  upon  the  other  at  the  place  of  contact  a 
force  whose  magnitude  increases  from  zero  up  to  some  maximum 
value  and  then  decreases  to  zero  again.  The  time  of  action  of  these 
forces  is  so  small  that  their  magnitudes  cannot  be  measured  ;  there- 
fore in. dealing  with  cases  of  impact  the  discussion  must  usually  be 
limited  to  a  consideration  of  the  total  changes  of  velocity  and  mo- 
mentum produced  in  the  bodies. 

Direct  and  indirect  impact. —  If  two  bodies  moving  in  the  same 
straight  line  collide,  their  impact  is  said  to  be  direct ;  their  original 
velocities  may  have  the  same  direction  or  opposite  directions.  If 
the  paths  of  the  two  bodies  before  collision  are  intersecting  lines, 
the  impact  is  indirect.  The  present  discussion  will  be  restricted 
to  the  case  of  direct  impact.  It  will  further  be  assumed  that  the 
bodies  are  of  such  symmetrical  form  that  the  forces  acting  be- 
tween them  when  in  contact  are  directed  along  their  common  line 
of  motion. 

Elastic  and  inelastic  bodies. —  As  the  two  bodies  come  into  col- 
lision, each  is  distorted  by  the  pressure,  the  amount  of  distortion  in- 
creasing as  the  pressure  increases.  If  the  bodies  are  wholly  inelastic, 
the  distortion  is  permanent,  the  bodies  showing  no  tendency  to  re- 
gain their  original  forms.  In  this  case  the  two  bodies  will  move  on 
with  a  common  velocity  after  the  impact.  If  they  are  elastic,  each 
tends  to  regain  its  original  form,  and  in  so  doing  exerts  a  force  upon 
the  other  so  that  the  two  tend  to  separate.  Elasticity,  as  thus  de- 
fined, is  possessed  by  bodies  in  various  degrees.  The  method  of 
specifying  the  degree  of  elasticity  will  be  considered  below. 


MOMENTUM    AND    IMPULSE.  283 

323.  Law  of  Action  and   Reaction  Applied  to  Impact. —  HA 

and  B  are  any  two  bodies  in  collision,  the  forces  exerted  by  them 
upon  each  other  are  at  every  instant  equal  and  opposite,  in  accord- 
ance with  Newton's  third  law  (Art.  259).  It  follows  that  the  impulses 
of  these  forces  during  any  time  are  equal  and  opposite.  Also,  since 
the  momentum  due  to  an  impulse  is  numerically  equal  to  the  impulse, 
the  two  bodies  receive  during  any  part  of  the  time  of  the  collision 
equal  and  opposite  quantities  of  momentum.  We  thus  reach  the  fol- 
lowing principle : 

The  collision  of  two  bodies  causes  no  change  in  their  total  mo- 
mentum. 

In  the  case  of  direct  impact  this  principle  may  be  stated  algebra- 
ically as  follows  :  Let  mx ,  m2  be  the  masses  of  two  bodies  moving  in 
the  same  straight  line,  vx ,  v2  their  velocities  before  collision  and  ©•/, 
V^  their  velocities  at  any  instant  during  or  after  the  collision.  (In 
any  particular  case  algebraic  signs  must  be  given  to  the  velocities  in 
accordance  with  their  directions  along  the  line  of  motion.)  Before 
the  collision  the  total  momentum  of  the  two  bodies  is 

mlvl  -f  m,v2 ; 
at  a  succeeding  instant  its  value  is 

mxvx  -f-  m2v2. 
By  the  above  principle  these  two  quantities  are  equal,  giving  the 

"  mxvx  -f-  m2v2  =  mxvx  +  m2v2.  .         .     (1) 

Equation  (1)  is  not  sufficient  for  the  determination  of  vx  and  v2 ; 
but  if  an  additional  relation  between  these  quantities  is  known  they 
may  be  determined.  For  example,  suppose  there  is  some  instant 
during  the  collision  at  which  the  two  bodies  have  the  same  velocity 
v ;  then  for  that  instant  vx  =  v2  =  v,  and  (1)  becomes 

mxvx  -f  m2v2  —  (mx  -f  #z2)^> 
or  v  =  (mxvx  -f  m2v2)/(mx  +  m2).  .         .     (2) 

324.  Inelastic  Impact. —  If  two  bodies  moving  in  the  same  line 
are  inelastic,  they  will  not  separate  after  collision  (Art.  322)  ;  hence 
in  this  case  equation  (2)  will  give  their  common  velocity  after  im- 
pact. 

Examples. 

1.  A  sphere  of  mass  10  lbs.  moving  at  the  rate  of  20  ft.-per-sec. 
overtakes  a  sphere  of  mass  20  lbs.  moving  at  the  rate  of  10  it.  -per- 


284  THEORETICAL    MECHANICS. 

sec.      If  the  bodies  are  inelastic,  what  is  their  velocity  after  the  col- 
lision? Ans.    13^  ft.-per-sec. 

2.  In  Ex.  1,  what  is  the  value  of  the  impulse  acting  on  each  body 
during  the  collision? 

Taking  the  direction  in  which  the  bodies  are  moving  as  positive, 
the  momentum  of  the  first  body  changes  from  200  to  133/^  (F.  P.  S. 
units),  while  the  momentum  of  the  second  changes  from  200  to 
266^3.  Hence  the  impulse  acting  upon  the  first  body  is  — 66^3 
poundal-seconds,  and  that  acting  on  the  second  is  +  66^3  poundal- 
seconds. 

3.  A  body  whose  mass  is  5  kilogr. ,  moving  at  the  rate  of  700 
cm.  -per-sec. ,  meets  a  body  whose  mass  is  4  kilogr.  Both  bodies  are 
brought  to  rest  by  the  collision.  Required  (a)  the  original  velocity 
of  the  second  body,  and  (b)  the  value  of  the  impulse  acting  on  each 
body  during  the  collision. 

Ans.  (#)  — 875  cm. -per-sec.  (J?)  Two  opposite  impulses  each 
equal  to  3. 5  X  io6  dyne-seconds. 

4.  Two  bodies  collide  while  moving  in  opposite  directions  with 
velocities  inversely  proportional  to  their  masses.  Show  that  their 
total  momentum  is  zero  at  every  instant  during  the  collision.  If  they 
are  inelastic,  what  is  their  common  velocity  at  the  end  of  the  col- 
lision ? 

5.  A  body  of  2  lbs.  strikes  a  body  of  50  lbs.  which  is  at  rest  on 
a  smooth  horizontal  surface.  Immediately  after  the  collision  the 
heavier  body  has  a  velocity  of  4  ft.  -per-sec.  while  the  lighter  body  is 
at  rest.  What  was  the  initial  velocity  of  the  lighter  body  ?  What 
total  impulse  acted  on  each  body? 

6.  Is  the  solution  of  Ex.  5  changed  by  assuming  the  horizontal 
surface  to  be  rough?     [See  Art.  321.] 

7.  A  2-oz.  bullet  passes  through  a  block  of  wood  weighing  4 
lbs.,  its  velocity  being  thereby  changed  from  1,100  ft.-per-sec  to 
950  ft.-per-sec.  With  what  velocity  does  the  block  move  after  the 
impact,  if  free?  What  is. the  value  of  the  total  impulse  acting  on 
the  block  ? 

8.  A  block  weighing  10  lbs.  is  struck  by  a  2-oz.  bullet  which 
remains  imbedded  in  it.  After  the  collision  the  block  has  a  velocity 
of  10  ft.-per-sec.     What  was  the  original  velocity  of  the  bullet? 

A  ns.  8 1  o  ft.  -per-sec 

325.  Elastic  Impact. —  If  two  elastic  bodies  moving  in  the  same 
straight  line  collide,  it  is  convenient  to  divide  the  time  occupied  by 
the  impact  into  two  parts  :  the  time  Tx  up  to  the  instant  of  greatest 
distortion  of  the  bodies,  and  the  time  T2  after  that  instant.  At  the 
instant  of  greatest  distortion  the  bodies  have  the  same  velocity. 
Before   that  instant  the  forces  acting  between   them   are   resisting 


MOMENTUM    AND    IMPULSE.  285 

their  approach,  after  that  instant  the  forces  are  urging  their  sepa- 
ration. 

Let  the  whole  impulse  which  either  body  exerts  upon  the  other 
be  divided  into  two  parts  ;  the  part  acting  during  Tx  being  called  the 
impulse  of  compression,  and  the  part  acting  during  T2  the  impulse  of 
restitution.  Let  the  ratio  of  the  latter  impulse  to  the  former  be  called 
the  coefficient  of  restitution,  and  be  represented  by  e. 

Let  mx  and  m2  be  the  masses  of  the  bodies ;  vx  and  v2  their  ve- 
locities before  collision  ;  v  their  common  velocity  at  the  instant  of 
greatest  compression  ;  vXi  v2  their  velocities  after  the  collision.  Let 
Q  denote  the  impulse  acting  upon  mx  during  compression,  and  emx 
the  impulse  acting  on  the  same  body  during  restitution.  The  im- 
pulses acting  on  m2  are  then  —  Q  and  —  eQ. 

Consider  either  body,  as  that  whose  mass  is  mx .  Before  the  col- 
lision its  momentum  is  mxvx ;  at  the  instant  of  greatest  compression 
it  is  tnxv  ;  at  the  end  of  the  collision  it  is  mxvx.     Hence 

(increase  of  momentum  due  to  impulse    Q)  =  mx(v  —  vx)  ; 
(      "        "  "  "  "        eQ)  =  mx(yx' —  v). 

Or,  since  the  impulse  is  numerically  equal  to  the  momentum  it 
produces, 

Q  =  mx{v  —  V1);.         .         .         .     (3) 

eQ  =  mx(vx  —  v) (4) 

Eliminating  Q  and  solving  for  vx\ 

vx  =  (1  -\~  e)v  —  evx. 

But  by  equation  (2)  (Art.  323), 

v  =  (mxvx  +  m2v2)/(mx  -f  m2) ; 

which  substituted  in  the  value  of  vx  gives,  after  reducing, 

111 

vi  =  vx  -     — -a—  (1  +  «)(»-,  -  »,)•    •        •    (5) 

mx  -f-  m2 

In  like  manner,  by  writing  equations  similar  to  (3)  and  (4)  for 
the  other  body,  we  may  find 

<  =  *. =t-(i  +"')(«,  —  ».)•      •         •     (6) 

mx  -f-  m. 

Equations  (5)  and  (6)  determine  the  velocities  of  the  bodies  after 
they  separate.      The  value  of  the  impulse  Q  and  of  the  total  impulse 


286  THEORETICAL    MECHANICS. 

(i  -j-  e)Q  may  be  determined  from  the  equation  of  impulse  and  mo- 
mentum. Thus,  the  total  change  of  momentum  of  the  body  mx  is 
mx(y{  —  vx).  Equating  this  to  the  total  impulse,  and  using  the  value 
of  vx  given  by  equation  (5), 

(1  +e)Q  =  -       m<m<    (i  +  *)0,  -v2);  .     (7) 

Q  = '—t—ivr  —  v,).       .         .         .     (8) 

Equation  (7)  gives  the  whole  impulse  acting  upon  the  body  ml. 
Assuming  vx  greater  than  v2 ,  the  impulse  is  negative.  Equation  (8) 
states  that  the  impulse  up  to  the  instant  of  greatest  compression  is 
independent  of  e,  that  is,  independent  of  the  degree  of  elasticity. 

The  value  of  e  may  lie  anywhere  between  o  and  1 .  For  inelastic 
impact  e  =  o,  and  the  value  of  vx  is 

vx  =  {inxvx  -f  m2v2)j(mx  -f  m2)}  .         .     (9) 

as  in  Art.  324.     For  perfect  restitution,  e  =  1,  and 

ITU 

v;  =  vx -*—&,—  v2)\      .      .    (10) 

2nt\       r  \  r      \ 

v2  =  v% -±—  (y2  —  vx).  .        .(11) 

mx  -\-  mt 

Examples. 

1.  A  man  weighing  160  lbs.  leaped  into  a  boat  whose  mass  was 
100  lbs.,  thereby  causing  it  to  move  from  rest  with  a  velocity  of  10 
ft.  -per-sec.     With  what  velocity  did  the  man  leap  ? 

An s.  16^  ft. -per-sec. 

2.  A  boat  weighing  200  lbs.  is  at  rest  in  still  water  when  a  person 
weighing  150  lbs.  starts  to  move  from  one  end  to  the  other.  If  he 
moves  10  ft.  relatively  to  the  boat,  how  far  does  he  move  relatively 
to  the  water,  and  how  far  does  the  boat  move  ? 

By  Newton's  third  law,  the  force  urging  the  man  forward  is  at 
every  instant  equal  to  the  force  urging  the  boat  in  the  opposite  di- 
rection. Let  P  be  the  magnitude  of  each  of  these  forces  in  poundals  ; 
then  at  any  instant  the  acceleration  of  the  boat  is  P/200  and  that  of 
the  man  P/i^o  in  the  opposite  direction.  The  velocities  due  to 
these  accelerations  in  any  time  are  in  the  same  ratio  as  the  accelera- 
tions ;  and  the  distances  described  by  the  two  bodies  in  the  same 
time  are  proportional  to  their  velocities,  and  therefore  to  their  accel- 
erations.    The  boat  thus  moves  three-fourths  as  far  as  the  man. 


MOMENTUM    AND    IMPULSE.  287 

3.  Two  perfectly  elastic  bodies  moving  along  the  same  line  col- 
lide.     Prove  that  they  exchange  velocities. 

4.  A  body  whose  mass  is  5  kilogr. ,  moving  at  the  rate  of  5 
met.  -per-sec. ,  collides  with  a  body  whose  mass  is  4  kilogr. ,  moving 
in  the  same  direction  at  the  rate  of  4  met.  -per-sec.  The  coefficient 
of  restitution  is  0.5.  Determine  the  velocities  of  the  bodies  after 
they  separate.  Ans.  4*/3  met. -per-sec.  and  4%  met. -per-sec. 

5.   In  Ex.  4,  compute  the  value  of  the  impulse  acting  during  com- 
pression, and  of  the  total  impulse  during  the  collision. 

Ans.  (2/9)  X  1  o6  dyne-seconds;  (1/3)  X  io6  dyne-seconds. 

6.  A  body  weighing  10  lbs.,  moving  at  the  rate  of  12  ft. -per-sec. , 
overtakes  a  body  weighing  18  lbs.,  moving  at  the  rate  of  5  ft. -per- 
sec.  After  the  collision  the  velocity  of  the  former  body  is  6  ft.  -per- 
sec.  Required  (a)  the  final  velocity  of  the  second  body ;  (b)  the 
coefficient  of  restitution.  Ans.  (a)  8^  ft. -per-sec.     (b)  }i. 

7.  Prove  that,  in  any  case  of  direct  collision,  the  ratio  of  the  rel- 
ative velocity  of  the  two  bodies  after  impact  to  their  relative  velocity 
before  impact  is  numerically  equal  to  the  coefficient  of  restitution. 

8.  A  rifle  weighing  3  lbs.  is  discharged  while  lying  on  a  smooth 
horizontal  plane.  The  weight  of  the  bullet  is  2  oz.  and  it  leaves 
the  barrel  with  a  velocity  of  1,400  ft. -per-sec.  What  will  be  the 
velocity  of  the  "  kick  "?     What  is  the  impulse  of  the  kick  ? 

9.  If  the  discharge  of  a  rifle  were  strictly  instantaneous,  what 
would  be  the  force  of  the  kick  ? 

10.  A  man  weighing  165  lbs.  leaps  from  a  boat  weighing  124 
lbs.  into  a  boat  weighing  150  lbs.  If  the  boats  are  initially  at  rest, 
compare  their  velocities  after  the  leap. 


§  2.   Motion  in  Any  Path. 

326.  Momentum  a  Vector  Quantity.  —  The  momentum  of  a 
moving  particle  has  been  defined  (Art.  312)  as  a  quantity  propor- 
tional directly  to  its  mass  and  to  its  velocity.  When  motion  in  a 
curved  path  is  studied,  it  must  be  remembered  that  direction  as  well 
as  magnitude  is  an  essential  part  of  the  value  of  velocity,  and  there- 
fore of  momentum.  Momentum  is,  in  fact,  a  vector  quantity,  its 
direction  coinciding  at  every  instant  with  that  of  the  velocity  of  the 
particle. 

327.  Moment  of   Momentum   or  Angular   Momentum. —  The 

momentum  of  a  particle  is  at  every  instant  associated  with  a  certain 
position.  It  is,  in  fact,  a  localized  vector  quantity  (Art.  26),  its  posi- 
tion-line at  any  instant  being  the  straight  line  along  which  the  motion 


288  THEORETICAL    MECHANICS. 

is   directed.     The  position-line   is   the   tangent  to  the  path   at  the 
instantaneous  position  of  the  particle. 

The  moment  of  momentum  of  a  particle  with  respect  to  any  point 
in  the  plane  of  the  motion  is  the  product  of  the  momentum  into  the 
perpendicular  distance  of  its  position-line  from  the  origin  of  moments. 
Moment  of  momentum  is  also  called  angular  momentum. 

328.  Increment  of  Momentum. —  If  the  velocity  of  a  particle 
varies  (either  in  direction  or  in  magnitude  or  in  both)  so  also  does 
the  momentum.     The  increment  of  momentum  for  any  interval  may 

be  found  in  the  manner  already  de- 
scribed for  finding  increment  of  velocity 
(Art.  246)  or  of  any  variable  vector 
(Art.  22).  Thus,  let  AB  (Fig.  142)  be 
the  path  described  by  a  particle  during 
any  interval,  vx  being  the  magnitude  of 
the  velocity  in  the  position  A  and  v2  its 
Fig.  T42.  magnitude  in  the  position   B.      At  the 

beginning  of  the  interval  the  momentum 
has  the  value  mi\ ,  directed  along  the  tangent  to  the  path  at  A  ;  at 
the  end  of  the  interval  its  value  is  mv% ,  directed  along  the  tangent  at 
B.  Representing  these  values  by  vectors  OA'  and  OB\  the  vector 
A'B'  represents  the  increment  of  momentum  for  the  given  interval. 

329.  Acceleration  of  Momentum  a  Vector  Quantity. —  The  ac- 
celeration of  momentum  (or  mass -acceleration)  of  a  particle  is  a 
vector  quantity  proportional  directly  to  the  mass  and  to  the  accel- 
eration. Its  direction  is  that  of  the  acceleration  ;  it  does  not,  in 
general,  coincide  with  that  of  the  tangent  to  the  path  at  the  position 
of  the  particle. 

Acceleration  of  momentum  bears  the  same  relation  to  momentum 
that  acceleration  bears  to  velocity.  It  may  be  defined  concisely  as 
momentum-increment  per  unit  time;  just  as  acceleration  is  defined  as 
velocity-increment  per  unit  time  (Art.  247). 

Examples. 

1.  If  the  velocity  of  a  particle  of  mass  18  lbs.  changes  from  12 
ft.-per-sec.  in  a  certain  direction  to  20  ft.-per-sec.  in  the  opposite 
direction,  what  is  the  momentum-increment? 

Ans.  576  F.  P.  S,  units. 


MOMENTUM    AND    IMPULSE.  289 

2.  If  the  velocity  of  a  particle  of  18  lbs.  mass  changes  from  12 
ft.-per-sec.  in  a  horizontal  direction  to  20  ft. -per-sec.  vertically  down- 
ward, what  are  the  magnitude  and  direction  of  the  momentum-incre- 
ment? Ans.  419.9  F.  P.  S.  units  ;  300  58'  from  vertical. 

3.  A  particle  of  60  lbs.  mass  describes  a  circle  of  6  ft.  radius  at 
the  uniform  rate  of  120  revolutions  per  minute.  Determine  the 
magnitude  and  direction  of  the  increment  of  momentum  received  by 
the  particle  while  making  (a)  one-sixth  of  a  revolution,  (b)  one-fourth 
of  a  revolution,  and  (c)  one-half  a  revolution. 

Ans.  (a)  4,524  F.  P.  S.  units. 

4.  A  particle  of  20  gr.  mass  describes  a  circle  of  90  cm.  radius 
with  a  uniform  speed  of  250  cm. -per-sec.  Determine  completely 
(in  magnitude  and  direction)  fhe  increment  of  momentum  in  one- 
hundredth  of  a  second. 

Ans.  Its  magnitude  is  139  C.  G.  S.  units,  nearly. 

5.  With  the  data  of  Ex.  4,  compute  the  moment  of  the  momentum 
{a)  about  the  center  of  the  circle ;  {b)  about  a  point  in  the  circum- 
ference, (c)  Compute  the  greatest  and  least  values  of  the  moment 
of  the  momentum  about  a  point  40  c  m.  from  the  center. 

330.  Impulse  of  a  Force  Whose  Direction  Varies. —  If  the 
direction  of  a  force  varies,  this  variation  must  be  taken  into  account 
in  computing  the  impulse.  Impulse  must  now  be  regarded  as  a 
vector  quantity,  its  direction  being  determined  by  that  of  the  force ; 
and  the  impulses  for  successive  intervals  of  time  are  to  be  combined 
by  vector  addition.  Thus,  if  Px'  and  P2f  denote  the  values  of  the 
impulse  for  two  successive  intervals  Atx  and  At.i}  the  impulse  for  the 
entire  interval  Atx  -f  At2  is  the  vector  sum  of  Px  and  P2f.  If  the 
force  has  a  constant  direction  during  the  interval  Atx ,  and  a  constant 
but  different  direction  during  the  interval  At.,,  Px  and  P.2'  may  each 
be  computed  as  in  Art.  315  or  Art.  316.  But  if  the  direction  of  the 
force  varies  continuously,  a  different  method  must  be  employed. 

Let  it  be  required  to  compute  the  impulse  of  any  variable  force 
P during  the  interval  from  t'  to  t" .  Let  the  interval  be  subdivided 
into  small  parts  Atx ,  At2 ,  .  .  .  ,  and  let  the  values  of  P  at  the 
beginnings  of  these  intervals  be  Px ,  P., ,  .  .  .  .  Approximate 
values  of  the  impulses  for  the  successive  small  intervals  are 

PxAtx      in  direction  of  Plf 

P.At2      in  direction  of  Ps , 


and   the   required   total   impulse  P'  is   approximately  equal  to   the 
*9 


29O  THEORETICAL    MECHANICS. 

vector  sum  of  /^A/, ,  P2At, ,  .  .  .  .  The  approximation  may 
be  made  as  close  as  desired  by  taking  the  intervals  sufficiently  small. 
The  exact  value  of  the  total  impulse  is  the  limit  approached  by  the 
approximate  value  as  A/, ,  A/2       ...     all  approach  zero.     That  is, 

P'  =  limit  IP  Ah  ~h  PA*t  +     •      •     •      ] . 
it  being  understood  that  the  -f-  sign  denotes  vector  addition.* 

It  will  be  noticed  that  the  above  process  of  computing  im- 
pulse is  independent  of  the  path  of  the  particle  upon  which  the  force 
acts.  So  far  as  its  magnitude  and  direction  are  concerned,  impulse 
depends  only  upon  force  and  time. 

331.  Vector  Equation  of  Impulse  and  Momentum. — The  prin- 
ciple of  impulse  and  momentum,  already  stated  for  the  case  of 
rectilinear  motion,  may  now  be  extended  to  the  case  of  a  particle  de- 
scribing any  path  under  the  action  of  a  force  varying  in  any  manner. 

The  impulse  of  t lie  force  during  any  interval  is  equal  in  magni- 
tude and  direction  to  the  change  of  momentum  of  the  particle  during 
that  interval. 

This  statement  includes  the  proposition  given  in  Art.  317.     The 

equation  .  , 

n  impulse  =  momentum-increment 

must  now  be  interpreted  as  a  vector  equation  ;  in  the  special  case  of 
rectilinear  motion  it  reduces  to  an  algebraic  equation. 

332.  Algebraic  Computation  of  Impulse  of  Variable  Force. — 
Usually  the  simplest  method  of  computing,  the  value  of  the  impulse 
of  a  force  whose  direction  varies  is  to  resolve  the  force  into  compo- 
nents parallel  to  fixed  rectangular  axes  and  determine  the  impulse  of 
each  axial  component  separately.  Restricting  the  discussion  to  the 
case  in  which  the  path  of  the  particle  and  the  line  of  action  of  the 
force  lie  in  a  plane,  let  P  be  the  value  of  the  force  at  time  /,  and  let 
the  axial  components  of  P  be  X  and  Y.  Let  P'  denote  the  impulse 
of  Pfor  the  interval  from  t'  to  /",  X'  and  Y'  being  the  axial  com- 
ponents of  P' .     Then 

X'  =  f  Xdt;      V  =  C    Ydt. 


.  *  The  above  reasoning  may  be  expressed  in  the  language  of  differentials, 
as  follows :  Let  /'be  a  vector  symbol  denoting  the  value  of  the  force  at  the 
instant  I,  and  P/  a  vector  symbol  denoting  the  value  of  the  required  impulse. 
Then  dP/  =  P  dt.  This  is  a  differential  equation,  both  members  of  which  are 
vectors. 


MOMENTUM    AND    IMPULSE.  29 1 

From  the  values  of  X!  and  V  the  magnitude  and  direction  of  P' 
may  be  determined. 

333.  Algebraic   Equations  of   Impulse   and   Momentum. —  In 

the  case  of  plane  motion  the  vector  equation  of  impulse  and  mo- 
mentum is  equivalent  to  two  algebraic  equations.  These  may  be 
deduced  immediately  from  the  differential  equations  of  motion  (Art. 
287).     These  equations  are 

X  =  mXy      Y  =z=  my. 

Multiplying  each  through  by  dt  and  integrating  between  limits  f 
and  /",  the  values  of  x  and  y  at  the  limits  being  x ',  x"  and  y' ,  y" 
respectively, 


\     Xdt=  m(x"  —  x')\      f    Vdt  =  m(y"—y' 


The  first  members  of  these  equations  are  the  axial  components  of  the 
impulse  of  the  resultant  force  acting  upon  the  particle  (Art.  332); 
the  second  members  are  the  axial  components  of  the  increment  of 
momentum  of  the  particle.  Together  the  equations  express  the 
proposition  that  the  resultant  impulse  is  equal  in  magnitude  and 
direction  to  the  total  increment  of  momentum.  The  general  principle 
of  Art.  331  is  thus  an  immediate  consequence  of  the  fundamental 
equations  of  motion  of  a  particle. 

334.  Impulse  and  Momentum  as  Localized  Vector  Quanti- 
ties.—  In  the  foregoing  discussion  no  reference  has  been  made  to  the 
path  of  the  particle  nor  to  the  position  in  space  of  the  line  of  action 
of  the  force.  These  must,  however,  be  taken  into  account  in  a  com- 
plete explanation  of  the  principle  of  impulse  and  momentum. 

The  momentum  of  a  particle  is  at  every  instant  directed  along  a 
definite  line  in  space.  This  may  be  regarded  as  its  position-line, 
momentum  being  thus  a  localized  vector  quantity  (Art.  26). 

A  force  acting  upon  a  particle  has  at  every  instant  a  definite  line 
of  action.  The  impulse  of  a  force  whose  line  of  action  remains 
constant  may  therefore  be  regarded  as  a  localized  vector  quantity 
whose  position-line  is  the  line  of  action  of  the  force.  If  the  line  of 
action  of  the  force  varies,  the  elementary  impulse  Pdt  correspond- 
ing to  an  elementary  time  dt  may  be  regarded  as  a  localized  vector 
quantity  whose  position-line  coincides  with  the  instantaneous  position 
of  the  line  of  action  of  the  force. 


292  THEORETICAL    MECHANICS. 

335.  Angular  Impulse. —  Let  P  denote  the  value  of  a  force  at 
any  instant  /,  and  P'  the    impulse    during   the    time  from  t'  to  /". 
Choosing  any  origin  of  moments  0  (Fig.  143),  let  a  be  the  perpen- 
dicular distance  from  O  to  the  line  of  ac- 
tion P.     Then  Pa  is  the   moment  of  the 
force  with  respect  to  0.      During  a  small 
interval  At  let  the  particle   to  which  the 
force  is  applied  move  a  short  distance  AB. 
The  change  in  P  will  be  small  ;  the  impulse 
during  the  small  interval  will  be  approxi- 
mately equal  to   PAt ;  and  the  moment  of 
q  '  the  impulse  will  differ  little  from  Pa  At. 

Fig.  143.  Let  the  whole  time  from  t'   to  t"   be 

subdivided  into  small  parts  A^, ,  A/2 ,  .   .   .  , 

and  let  the  values  of  P  at  the  beginnings  of  these  small  intervals  be 

Plt  P2i      .     .      .      ,  the  corresponding  values  of  the  arm  a  being  ax , 

a „     .     .     .     .     The  sum 

PxaxAtx  +  P2a2At,  -f     .     .     . 

may  be  regarded  as  an  approximate  value  of  the  moment  of  the 
resultant  impulse  P .  Taking  the  intervals  smaller  and  smaller, 
approaching  zero,  the  exact  value  of  the  moment  of  the  impulse  is 

limit  \\PxaxAtx  +  P,a2At,  -f     .     .     .     ]  =  f  Padt. 

If  the  moment  of  the  force  with  respect  to  O  is  represented  by  G, 

G  =  Pa, 
and  the  moment  of  the  impulse  is  equal  to 


x 


Gdt. 


The  moment  of  the  impulse  of  a  force  about  a  given  point  is  also 
called  its  angular  impulse  about  that  point. 

336.  Position-Line  of  Resultant  Impulse. —  It  has  been  pointed 
out  that  the  impulse  of  a  force  during  an  elementary  time  dt  may  be 
regarded  as  having  a  definite  position-line,  even  when  the  line  of 
action  of  the  force  is  variable.  The  resultant  impulse  for  an  interval 
of  any  length  may  be  regarded  as  localized  in  a  line  whose  position 


P'a'  =  \     Padt. 


MOMENTUM    AND    IMPULSE.  293 

is  determined  by  the  value  of  the  moment  of  the  impulse  about  any 
given  point.  Its  distance  a!  from  the  origin  of  moments  must  satisfy 
the  equation 

f 

It  will  be  observed  that  the  method  of  determining  the  magni- 
tude, direction  and  position-line  of  the  resultant  impulse  for  any 
interval  is  essentially  the  same  as  that  of  finding  the  magnitude, 
direction  and  line  of  action  of  the  resultant  of  any  system  of  forces 
acting  in  the  same  plane  on  a  rigid  body  (Art.  97).  In  the  case  of 
forces,  the  resultant  is  equal  to  the  vector  sum  of  the  components ; 
and  the  moment  of  the  resultant  is  equal  to  the  algebraic  sum  of 
the  moments  of  the  components.  In  the  case  of  the  impulse,  the 
resultant  impulse  for  any  interval  is  equal  to  the  vector  sum  of  the 
impulses  for  the  elementary  intervals  ;  and  the  moment  of  the  re- 
sultant impulse  is  equal  to  the  algebraic  sum  of  the  moments  of  the 
elementary  impulses.  The  similarity  of  these  relations  may  be  fur- 
ther illustrated. 

Let  AB  (Fig.  144)  represent  the  path  described  by  the  particle 
during  the  time  from  t'  to  /".  Let  the  whole  time  be  subdivided 
into  small  intervals  A^ ,  A/2,  .  .  .  ;  let  A,  a,  b,  .  .  .  be 
the  positions  of  the  particle  at  the  beginnings  of  these  intervals  ;  and 
let  Px,  P2,  .  .  .  be  the  corre- 
sponding values  of  the  force,  their  di- 
rections being  as  shown  in  the  figure 
The  impulses  during  A/x ,  A/, ,  .  .  . 
have  approximately  the  values 

Px  At,     along  the  line  AA\ 

P2At2     along  the  line  aa\ 


Let    these   be   combined    as    if  they 
were  forces ;    their  resultant  is  equal 

to  their  vector  sum  and  acts  along  some  determinate  line,  as  M'N' ; 
and  this  resultant  is  an  approximate  value  of  the  resultant  impulse 
P'.  The  exact  value  is  the  limit  approached  by  the  approximate 
value  as  A/, ,  A/2 ,  .  .  .  are  all  made  to  approach  zero.  The  line 
M'N'  approaches  a  limiting  position  MN,  which  is  the  position-line 
of  the  resultant  impulse  P'. 


294 


THEORETICAL    MECHANICS. 


Fig.  145. 


Examples. 

1.  A  particle  is  projected  horizontally  with  a  given  velocity  Vt 
after  which  it  is  acted  upon  by  no  force  except  gravity.  Determine 
the  position-line  of  the  resultant  impulse  of  the  force  of  gravity 
during  any  time. 

Let  the  particle  be  projected  from  the  point  A  (Fig.  145)  in  the 
direction  AB',  and  let  AB  be  the  path  described  during  the  time  T. 
Let  this  time  be  subdivided  into  a  number  of  small  intervals,  each 

equal  to  At,  and  let  A,  a,  b,  .  .  . 
be  the  positions  of  the  particle  at  the 
beginnings  of  these  intervals.  If  m  is 
the  mass,  the  force  is  equal  to  mg  and 
is  constant  in  direction.  Applying  the 
method  above  described,  the  required 
impulse  is  approximately  equal  to  the 
resultant  of  a  series  of  equal  impulses 
mgAt  acting  in  lines  AA' ,  aa' ,  bb', 
.  The  successive  distances  be- 
tween these  lines  are  equal,  since  the 
horizontal  velocity  is  constant  ;  hence 
the  position-line  of  the  resultant  is  midway  between  the  extreme 
lines  AA',  gg '.  The  limiting  position  of  this  line,  as  At  is  made  to 
approach  zero,  bisects  AB'.  The  magnitude  of  the  resultant  im- 
pulse is  obviously  mgT. 

2.  In  Ex.  1,  compute  the  moment  of  the  resultant  impulse  by 
integration,  and  thus  verify  the  above  result. 

337.  Equation  of  Angular  Impulse  and  Angular  Momentum. — 
The  impulse  of  a  force  and  the  increment  of  momentum  produced 
by  it,  during  any  element  of  time,  are  not  only  equal  in  magnitude 
and  direction  but  have  the  same  position-line.  From  this  it  follows 
that  their  moments  about  any  origin  are  equal.  And  since  this 
equality  holds  for  every  element  of  time,  it  holds  for  any  interval 
whatever.     That  is, 

The  angular  impulse  of  a  force  for  any  interval  of  time  is  equal 
to  the  angular  momentum  produced  by  the  force  during  that  time. 

This  equation  and  the  two  given  in  Art.  332  are  the  complete 
algebraic  expression  of  the  principle  of  impulse  and  momentum  as 
applied  to  the  motion  of  a  particle  in  a  plane.     To  summarize  : 

The  proposition  that  impulse  is  equal  to  change  of  momentum, 
in  its  full  meaning,  must  be  understood  to  express  equality  of  magni- 
tude, agreement  in  direction,  and  identity  of  position-lines.  In  case 
of  plane  motion,   this   is  completely  expressed    by  three  algebraic 


MOMENTUM    AND    IMPULSE.  295 

equations.      Two  of  these  may  be  obtained  by  resolving  in  any  two 
directions,  the  third  by  taking  moments  about  any  point. 

338.  Geometrical  Illustration  of  General  Equation  of  Impulse 
and  Momentum.— Let  the  curve  AB  (Fig.  146)  be  the  path  de- 
scribed by  a  particle  during  the  interval 
from  t'  to  t" .  Let  v'  and  v"  be  the  in- 
itial and  final  values  of  the  velocity. 
From  any  point  0  draw  vectors  OA', 
OB',  representing  the  initial  and  final 
values  of  the  momentum.  These  vec- 
tors are  parallel  to  the  tangents  to  the 
path  at  A  and  at  B  respectively,  and 
*OA'  =  mv' ,  OB'  =  mv" .  The  incre- 
ment of  momentum  m  Av  is  then  rep- 
resented by  the  vector  A ' B' .  To  apply  the  principle  of  impulse 
and  momentum,  the  position-line  of  the  momentum-increment  must 
be  determined. 

The  position-line  of  the  initial  momentum  is  the  tangent  to  AB 
at  A  ;  that  of  the  final  momentum  is  the  tangent  to  AB  at  B ;  the 
position-line  of  the  momentum-increment  therefore  passes  through 
C,  the  intersection  of  these  tangents. 

Let  P*  be  the  resultant  impulse  applied  to  the  particle  during  the 
interval ;  then  P'  =  m  A?',  and  its  position -line  is  CD,  parallel  to 
A'B'. 

In  the  foregoing  explanation  of  the  general  principle  of  impulse 
and  momentum,  it  has  been  assumed  that  each  of  these  quantities  is 
subject  to  the  laws  of  composition  and  resolution  which  apply  to 
forces  acting  upon  a  rigid  body.  As  applied  to  impulse,  these  laws 
have  been  explicitly  stated  in  Art.  336.  As  applied  to  momentum, 
it  is  assumed  that  the  increments  of  momentum  received  by  a  par- 
ticle during  successive  intervals  are  to  be  combined  as  if  they  were 
forces  acting  on  a  rigid  body  in  order  to  produce  the  momentum- 
increment  for  the  entire  interval. 

These  are  arbitrary  assumptions,  amounting  in  reality  to  defini- 
tions of  resultant  impulse  and  resultant  momentum.  They  are  ex- 
tremely useful  because  they  make  it  possible  to  state  very  concisely 
the  full  meaning  of  the  equation  of  impulse  and  momentum.  These 
definitions  are  especially  valuable  when  the  general  principle  is  ex- 
tended to  systems  of  particles  and  to  rigid  bodies. 


296 


THEORETICAL    MECHANICS. 


Examples. 

1.  In  Ex.  1,  Art.  336,  show  that  the  tangent  to  the  path  at  B  bi- 
sects AB' .  [Apply  the  general  principle  of  impulse  and  momentum, 
using  the  result  already  found  in  the  solution  of  the  example.  ] 

2.  If  A  and  B  are  any  two  points  in  the  path  of  a  projectile, 
show  by  the  principle  of  impulse  and  momentum  that  the  tangents 
at  A  and  B  and  the  vertical  line  bisecting  the  chord  AB  intersect  in 
a  point.  [The  position-line  of  the  resultant  impulse  may  be  located 
as  in  Ex.  1,  Art.  336.] 

3.  A  particle  describes  a  circle  at  a  uniform  rate.  Show  that  the 
resultant  impulse  applied  to  it  during  any  interval  passes  through 
the  center.  Show  also  that  its  position-line  bisects  the  path  described 
during  the  interval. 

» 
339.  Sudden  Impulse. —  If  a  particle  receives  a  sudden  impulse 

whose  direction  is  inclined  to  that  in  which  the  particle  is  mov- 
ing, the  direction  of  motion  is  suddenly 
changed.  The  relation  between  the  in- 
itial and  final  velocities  and  the  impulse 
is  expressed  by  the  vector  equation  of 
impulse  and  momentum,  irrespective  of 
the  time  occupied  by  the  impulse. 

During  the  time  of  action  of  a  sud- 
den impulse  the  particle  moves  over  a 
very  short  distance.  The  direction  of 
the  motion  may,  however,  change  by 
any  angle.  In  general,  therefore,  a  sud- 
den impulse  causes  a  rapid  curvature  of 

the  path.     Let  a  particle  describing  the  straight  path  LA  (Fig.  147) 

receive  a  blow  whose  total  impulse  is  rep- 
resented   by   the   vector  A'B'.     If  OA' 

represents  the  initial  momentum,  OB'  will 

represent  the  momentum  after  the  blow. 

The  path  after  the  blow  will  be  some  line 

BM,  parallel  to  OB'.     The  path  during 

the    impulse   will    be    some   curve   AB. 

The  shorter    the    time    occupied    by   the 

blow,  the  shorter  the  path  AB.     If  the 

impulse  were  instantaneous,  A    and  B  would  fall  together, 

Fig.  148. 


147. 


as   in 


momentum  and  impulse.  297 

Examples. 

1.  A  ball  whose  mass  is  5^  oz.  is  moving  horizontally  at  the 
rate  of  ioo  ft.-per-sec.  when  it  is  struck  by  a  bat  in  such  a  way  that 
immediately  after  the  blow  it  has  a  velocity  of  150  ft.-per-sec.  in  a 
direction  making  an  angle  of  300  upward  from  the  horizontal.  Re- 
quired the  value  of  the  impulse. 

Ans.  79.34  poundal-seconds,  inclined  i8°4  upward  from  the 
horizontal,  assuming  the  horizontal  velocity  to  be  reversed  in  direc- 
tion by  the  blow. 

2.  In  Ex.  1,  suppose  the  ball  to  receive  the  same  blow  when 
moving  at  an  angle  of  io°  downward  from  the  horizontal.  How  will 
it  move  immediately  after  the  blow  ? 

3.  In  the  same  case,  if  the  blow  occupies  0.01  sec,  what  is  the 
average  value  of  the  force?  What  will  be  the  effect  of  gravity  on  the 
motion  of  the  ball  during  the  blow  ? 

340.  Collision  of   Bodies  Moving  in   Different   Paths. —  The 

collision  of  bodies  moving  in  different  paths  presents  a  less  simple 
problem  than  that  treated  in  Arts.  322-325.  No  discussion  of  this 
problem  will  be  given  here,  except  to  point  out  the  application  of  the 
law  of  action  and  reaction. 

As  a  consequence  of  this  law,  the  total  impulses  acting  on  two 
bodies  during  their  collision  are  equal  and  opposite.  It  follows  that 
their  momentum-increments  are  equal  and  opposite.  The  vector 
sum  of  their  momenta  therefore  has  the  same  value  after  collision  as 
before.  Its  value  is,  in  fact,  the  same  at  every  instant  during  the 
collision. 

Examples. 

1.  A  body  of  5  lbs.  mass  strikes  a  fixed  surface  which  deflects  it 
200  and  changes  its  speed  from  25  ft.-per-sec.  to  10  ft.-per-sec. 
What  is  the  total  impulse  exerted  upon  the  body  during  the  collision? 

Ans.  79.8  poundal-seconds,  directed  at  angle  1670  38'  with  the 
original  motion. 

2.  A  ball  A  of  6  lbs.  mass  is  at  rest  on  a  horizontal  plane  when 
it  is  struck  by  a  ball  B  of  4  lbs.  mass  moving  at  the  rate  of  12  ft.- 
per-sec.  After  the  collision  A  is  moving  at  the  rate  of  4  ft.  -per-sec. 
in  a  direction  inclined  6o°  to  the  original  velocity  of  B.  How  does 
B  move  after  the  collision  ? 

Ans.  With  velocity  10. 4  ft. -per-sec.  at  right  angles  to  motion  of  A. 

3.  Two  bodies  moving  in  opposite  directions  with  velocities  in- 
versely proportional  to  their  masses  collide  in  such  a  way  as  to  be 
deflected  from  the  original  directions  of  motion.  Show  that  after  the 
collision  they  move  in  opposite  directions  with  velocities  inversely 
proportional  to  their  masses. 


CHAPTER   XVII. 

WORK  AND    ENERGY. 

§  i.    Work  in  Case  of  Rectilinear  Motion. 

341.  When  a  Force  Does  Work. —  A  force  is  said  to  do  work 
upon  the  body  to  which  it  is  applied  when  its  point  of  application 
moves  in  the  direction  toward  which  the  force  acts. 

342.  Quantity  of  Work  Done  by  Constant  Force. —  The  amount 
of  work  done  by  a  constant  force  is  a  quantity  proportional  directly 
to  the  magnitude  of  the  force  and  to  the  distance  moved  by  its  point 
of  application  in  the  direction  of  action  of  the  force.  The  numerical 
value  of  the  work  depends  upon  the  unit  in  which  it  is  expressed. 

The  unit  of  work  usually  chosen  is  the  quantity  of  work  done  by 
the  unit  force  when  its  point  of  application  moves  the  unit  distance 
in  the  direction  of  action  of  the  force.  If  the  pound  is  taken  as  the 
unit  force  and  the  foot  as  the  unit  distance,  the  unit  work  is  the 
quantity  of  work  done  by  a  force  of  one  pound  when  its  point  of 
application  moves  one  foot  in  the  direction  of  action  of  the  force. 
This  unit  is  called  a  foot-pound,  and  is  the  unit  commonly  employed 
by  engineers.  In  the  C.  G.  S.  system  the  unit  of  work  is  the  quan- 
tity of  work  done  by  a  force  of  one  dyne  while  its  point  of  application 
moves  one  centimeter  ;  the  name  erg  has  been  given  to  this  unit.* 
In  the  F.  P.  S.  system  the  unit  of  work  is  the  foot-poundal. 

When  the  point  of  application  of  a  force  moves  a  certain  distance 
in  the  direction  of  action  of  the  force,  the  force  is  said  to  act  through 
that  distance.  The  unit  work  may  therefore  be  defined  as  the  quan- 
tity of  work  done  by  the  unit  force  acting  through  the  unit  distance. 


*  The  erg  is  very  small  in  comparison  with  the  foot-poundal  or  the  foot- 
pound.    Thus, 

1  foot-poundal  =  421,390  ergs; 

1  foot-pound  —  1.356  X  io7  ergs. 
(The  last  result  assumes^  =  981  C.  G.  S.  units.) 

For  practical  use,  especially  in  electrical  engineering  applications,  an- 
other unit  of  work,  the  joule,  has  been  introduced,  defined  as  equal  to  io7 
ergs. 


WORK    AND    ENERGY.  299 

343.  Positive  and  Negative  Work. —  If  the  point  of  application 
of  a  force  moves  in  a  direction  opposite  to  that  of  the  force,  the  force 
is  said  to  do  negative  work  upon  the  body. 

If  P  is  the  algebraic  value  of  a  constant  force,  and  Ax  that  of  the 
displacement  of  its  point  of  application,  the  positive  direction  being 
the  same  for  both,  the  value  of  the  work  is  given  with  proper  sign  by 
the  product  p^x 

Examples. 

1.  A  body  weighing  10  lbs.  is  thrown  upward  against  gravity. 
Compute  the  work  done  upon  it  by  its  weight  (a)  while  it  rises  10 
ft.  and  (£)  while  it  falls  10  ft. 

Ans.   (a) — ioo^foot-poundals.      (b)  -\-ioog  foot-poundals. 

2.  If  the  resistance  of  the  air  amounts  to  a  constant  force  of  2  lbs., 
compute  the  work  done  by  it  in  both  cases  of  Ex.  1. 

Ans.  — 20  foot-pounds  in  each  case. 

344.  Work  Done  by  Variable  Force. —  If  the  value  of  a  force 
varies  during  the  displacement  of  its  point  of  application,  the  work 
must  be  computed  by  integration. 

Let  OB  (Fig.  149)  be  the  path  described  by  the  point  of  appli- 
cation (also  the  line  of  action  of  the  force).  Let  P  be  the  value  of 
the  force  when  the  point  of  application  is  at  distance  x  from  0,  and 
let  it  be  required  to  compute  the  work 

done  by  P  while  the  point  of  applica-        j- Xt-- h 

tion  moves  from  A  to  B.  ,►--••  x\ -."  •"•"•~~;j 

Divide  AB  into  small  parts  A  a       O  A  a  b  c  B 

-=  Axlt  afr  =  Ax2,  &C  =  Ax.,y     .     .     .  Fig.  149. 

and  let  Plt  Pt1  P,,      .      .      .     be  the 

values  of  P  when  the  point  of  application  is  at  A,  a,  b,      .     .     . 
Let  W  represent  the  required  work  ;  then,  approximately, 

W=  PxAxx  +  P2Axt  +  PA**  +     .... 

By  taking  A^Tj,  Ax.2,  .  .  .  small  enough,  the  approximation 
may  be  made  as  close  as  desired.  The  exact  value  of  the  work  is 
the  limit  approached  by  the  approximate  value  as  Axly  Ax.i} 
.  .  .  are  made  to  approach  zero.  That  is,  if  xi  =  OA  and 
x2  =  OB, 

W  =  limit  [/>  Axx  +  P,  Axt  ■+     .     .     .     ]  = f  Pdx. 

*i 


A 


3OO  THEORETICAL    MECHANICS. 

345.  Graphical  Representation  of  Work  Done  by  Variable 
Force. —  At  every  point  of  the  line  AB  representing  the  displace- 
ment of  the  point  of  application  (Fig. 
150),  erect  an  ordinate  whose  length 
is  equal  to  the  value  of  P  for  that 
position  ;  the  extremities  of  these  or- 
dinates  lie  on  a  curve  A'B'.  The 
0  j{  B~       area  bounded  by  the  curve,  the  line 

Fig.  150.  AB,  and  the  ordinates  AA',  BB\  is 

equal  to  the  work  done  by  P  during 
the  displacement  AB.  For,  the  expression  above  found  for  the 
work, 

Pdx, 


L 


is  also  the  value  of  the  area. 

Let  the  work-diagram  be  drawn  in  each  of  the  following  examples. 

Examples. 

1.  A  body  is  suspended  by  an  elastic  string  whose  unstretched 
length  is  4  ft.  Under  a  pull  of  10  lbs.  the  string  stretches  to  a  length 
of  5  ft.  Required  the  work  done  on  the  body  by  the  tension  of  the 
string  while  its  length  changes  from  6  ft.  to  4  ft. 

Ans.  20  foot-pounds. 
*    2.  A  body  whose  mass  is  m  falls  vertically  to  the  earth's  surface 
from  a  height  equal  to  the  radius  R.     Compute  the  work  done  by 
the  earth's  attraction  during  the  fall.        Ans.  \mgR  kinetic  units. 

3.  In  Ex.  2,  let  m  =  100  lbs.,  P  ==  21,000,000  ft.  Give  the 
value  of  the  work  in  foot-pounds.  Ans.   1.05  X   io9. 

4.  A  particle  moves  in  a  straight  line  under  the  action  of  a  force 
directed  toward  a  fixed  point  in  the  line  of  motion  and  varying  di- 
rectly as  the  distance  from  that  point.  Compute  the  work  done  on 
the  particle  during  a  given  displacement. 

Ans.  Let  P'  be  the  value  of  the  force  when  the  particle  is  at  the 
unit  distance  from  the  fixed  point,  xx  and  xt  the  initial  and  final  dis- 
tances from  that  point.       Then  the  required  work  is  \P'{x*  —  x*). 

346.  Dimensions  of  Unit  Work.  —  If  the  units  of  force  and  length 

are  both  chosen  arbitrarily,  the  unit  work  has  dimensions  FL.     But 

in  the  kinetic  system    (Art.  219),  F  =  ML/T2,  which  reduces  the 

unit  work  to  the  dimensions 

ML2/T2. 


WORK    AND    ENERGY. 


3OI 


§  2.    Work  in  Any  Motion  of  a  Particle. 

347.  General  Definition  of  Work. —  When  the  point  of  applica- 
tion of  a  force  receives  a  displacement  which  is  not  along  the  line  of 
action,  the  definition  of  work  must  be  enlarged. 

Let  a  force  P  remain  constant  in  magnitude  and  direction  while 
its  point  of  application  describes  any  straight  line  AB  (Fig.  151). 
The  quantity  of  work  done  depends  upon  the  value  of  AB' ,  the 
orthographic  projection  of  AB  upon  the  line  of  action  of  P.  This 
projection  may  be  called  the  effective  displacement. 

The  quantity  of  work  done  by  a  constant  force  while  its  point  of 
application  receives  any  rectilinear  displacement  is  equal  to  the  prod- 
uct of  the  force  into  the  effective  displacement. 

If  the  direction  of  the  effective  displacement  coincides  with  that 
of  the  force,  the  work  is  positive  ;  if  the  effective  displacement  is  op- 
posite to  the  force  the  work  is  negative.      If  the  actual  displacement 


Fig.  151. 


Fig.  152. 


Fig.  153. 


is  at  right  angles  to  the  direction  of  the  force,  the  effective  displace- 
ment is  zero.     These  three  cases  are  illustrated  in  Figs.  151,  152, 

153- 

The  cases  of  positive  and  negative  work  may  be  illustrated  by  a 
body  sliding  along  an  inclined  plane.  Let  P  be  the  weight  of  the 
body.  In  computing  the  work  done  by  the  force  P,  the  effective 
displacement  is  the  vertical  projection  of  the  distance  moved  by  the 
body.  If  the  body  descends  a  vertical  distance  //,  the  work  is  -Ph. 
If  it  rises  a  vertical  distance  h,  the  work  is  — Ph.  If  the  plane  is 
horizontal,  the  work  is  zero. 

348.  Work  Done  by  Constant  Force  Whose  Point  of  Applica- 
tion Receives  Any  Displacement.—  The  above  rule  for  computing 
the  work  of  a  constant  force  may  be  extended  to  the  case  in  which 
the  displacement  follows  any  curve  AB.  It  may  be  shown  that  the 
quantity  of  work  done  by  the  force  is  equal  to  the  product  of  the 


302  THEORETICAL    MECHANICS. 

magnitude  of  the  force  into  the  projection  of  AB  upon  aline  parallel 
to  the  force.  For,  suppose  the  path  to  be  replaced  by  a  broken  line 
AabB  (Fig.  154),  a  and  b  being  any  points  of  the  curve.  The  work 
done  during  the  displacement  along  this  broken  line  is  equal  to 

P{Aa'  +  a'b'  +  b'B')  =  PX  AB'. 

It  is  evident  that  the  same  value  of  the  work  results  if  the  displace- 
ment follows  any  broken  line  whatever  joining  A  and  B.  If  the 
d  number  of  points  such  as  a  and  b  is  taken 

greater  and  greater,  the  broken  line  ap- 
proaches the  curve  as  a  limit ;  hence  the 
work  done  during  the  motion  along  the 
curve  has  the  value  P  X  AB' . 
0      l?7^ — ■*  It  follows   that  the  work   done  by  a 

pIG   I54  force  which  remains   constant  in  magni- 

tude and  direction  while  its  point  of  ap- 
plication moves  from  A  to  B  has  the  same  value  whatever  the  path 
described  between  A  and  B.  An  example  of  this  is  the  work  done 
by  gravity  during  the  movement  of  bodies  near  the  earth's  surface. 

349.  General  Algebraic  Formula  for  Work  of  Constant  Force. — 
The  general  rule  for  computing  the  work  done  by  a  force  P  which 
remains  constant  in  magnitude  and  direction  while  its  point  of  appli- 
cation moves  from  A  to  B  along  any  path  may  be  stated  algebraic- 
ally in  the  form  n  ..   ,  A  D         a. 

J  P  X  (AB  cos  0), 

in  which  0  is  the  angle  between  the  straight  line  AB  and  the  direc- 
tion of  P.  The  work  is  positive  or  negative,  according  as  the  angle 
0  is  less  or  greater  than  900  (it  may  always  be  so  measured  as  not  to 
exceed  180°). 

Since  P  X  (AB  cos  6)  =  (P  cos  6)  X  AB,  it  is  evident  that 
the  work  done  is  equal  to  the  product  of  tlie  displacement  into  the 
resolved  part  of  the  force  in  the  direction  of  the  displacement.  It  is 
useful  to  remember  the  rule  in  both  forms. 

350.  Work  Done  by  Variable  Force  During  Any  Displace- 
ment of  Its  Point  of  Application. —  Let  a  force  P  vary  in  magni- 
tude and  direction  while  the  particle  to  which  it  is  applied  describes 
any  curved  path  AB  (Fig.  155).  On  AB  choose  any  number  of 
points  a,  b,  c,  .  .  .  ,  dividing  the  curve  into  small  parts.  Let 
Piy  P2,     .      •      •     be  the  values  of  the  magnitude  Pwhen  the  par- 


WORK    AND    ENERGY. 


303 


tide  is  at  A,  a,  .  .  .  ;  and  let  0{  =  angle  between  Px  and 
chord  Aa>  9.,  =  angle  between  P2  and  chord  ab,  etc.  Then  an 
approximate  value  of  the  work  done  by  the  force  is 

Px  cos  6X  {A a)  +  P%  cos  d,  (ab)  -f  P,  cos  03  (&)'+■'.'... 

The  smaller  the  parts  ^4#,  #$,  <fc,  etc.,  the  closer  the  approximation. 
The  true  value  of  the  work  is  the  limit  of  the  above  quantity  as  the 
number  of  points  a,  b,  c,  etc. ,  is  increased 
indefinitely.    That  is,  if  W  is  the  true  value 
of  the  work, 

W  =  limit  [(Px  cos  6X)  X  Aa  + 

(P2  cos  6,)  X  ^  +     .     .     •     ] 

=  f(P  cos  0>&. 

Here  0  represents  the  angle  between  the 
tangent  to  the  curve  at  any  point  and  the 

corresponding  direction  of  P,  s  is  the  length  of  the  curve  from  some 
fixed  point  to  the  position  of  the  particle,  and  the  limits  of  the  inte- 
gration must  be  so  taken  as  to  include  the  entire  displacement  AB. 
If  s '  and  s "  are  the  values  of  s  at  A  and  at  B  respectively, 


W=  C(P  cos  0)ds. 


351.  Work  of  Central  Force. —  If  the  force  is  always  directed 
toward  a  fixed  point,  the  above  expression  for  the  work  may  be  re- 
duced to  a  convenient  form  as  follows: 

Let  O  (Fig.  156)  be  the  center  of  force,  and  AB  the  path.  Let 
r  be  the  radius  vector  of  the  particle  measured  from  O,  rf  and  r" 
being  the  values  of  r  at  A  and  B.  Let  s  be 
measured  positively  in  the  direction  AB. 
Drawing  MN  tangent  to  the  path,  0  is  the 
angle  OMN.     Evidently, 

dr  —  ds  cos  (w  —  0)  =  —  ds  cos  6. 

The  above  value   of  the  work   therefore  re- 
duces to 


w-^=  —  fr 


Pdr 


If  P  is  a  known  function  of  /',  IV  may  be  found  by  direct  integration. 


304  THEORETICAL    MECHANICS. 

This  formula  shows  that  if  a  particle  moves  under  the  action  of  a 
central  force  which  is  any  function  of  the  distance  from  the  center, 
the  work  done  by  the  force  during  any  motion  depends  only  upon 
the  initial  and  final  distances  from  the  center. 

Examples. 

1.  Compute  the  work  done  on  a  particle  by  an  attractive  force 
varying  according  to  the  law  of  gravitation. 

Let  P  be  the  magnitude  of  the  force  at  any  distance  r  from  the 
center  of  attraction,  and  let  Px  be  the  value  of  P  when  r  =  rx.  Ac- 
cording to  the  law  of  gravitation  we  have 

P  :  p  =  r  ■  :  r'l\ 
or  P  =  PirJ/r*  =  kjr\ 

k  being  a  constant.      If  the  body  moves  so  that  the  initial  and  final 
values  of  r  are  r'  and  r",  the  work  done  is 


w 


rpdr=-C**=k[r--i\ 


2.  In  Ex.  1,  let  the  earth  be  the  attracting  body. 

In  this  case,  if  G  denotes  the  weight  of  the  body  at  the  earth's 
surface,  and  R  the  radius  of  the  earth,  we  have  k  =  GR2',  hence 
W=  GR2(i/r"  —  i/r'). 

3.  If  a  body  weighs  1,000  lbs.  at  the  earth's  surface,  how  much 
work  will  be  done  upon  it  by  the  earth's  attraction  while  it  falls  to 
the  surface  from  an  elevation  of  10,000,000  ft.?  (Take  R  = 
21,000,000  ft.) 

4.  A  particle  moves  under  the  action  of  a  central  force  varying 
directly  as  the  distance  from  the  center.  Compute  the  work  done 
by  the  force  during  any  motion. 

[Since  the  work  done  depends  only  upon  the  initial  and  final  dis- 
tances from  the  center  of  attraction,  it  is  evident  that  it  has  the  same 
value  as  in  Ex.  4,  Art.  345.] 

5.  A  body  describes  a  straight  line  while  acted  upon  by  a  force 
of  constant  magnitude  P  lbs.  directed  always  toward  a  fixed  point 
distant  h  ft.  from  the  line.  Determine  the  work  done  by  the  force 
while  the  body  passes  between  any  two  given  positions. 

6.  In  Ex.  5,  let  h  =  10  ft.,  P=  14  lbs.  Compute  the  work 
done  by  the  force  while  the  body  moves  40  ft. ,  starting  from  its  near- 
est position  to  the  fixed  point.  Ans.   — 437  foot-pounds. 

352.  Work  Done  by  Resultant  of  Any  Number  of  Forces. — 

The  quantity  of  work  done  by  the  resultant  of  any  number  of  con- 
current forces  during  any  displacement  of  their  point  of  application 


WORK    AND    ENERGY.  305 

is  equal  to  the  algebraic  sum  of  the  quantities  of  work  done  by  the 
several  forces  duri?ig  tJie  same  displaceme?it. 

Let  P  be  the  resultant ;  Pu  P2,  .  .  .  ,  the  several  forces  ; 
0t  dlt  0.it  .  .  .  ,  the  angles  between  P,  Px,  P2)  .  .  .  and 
the  tangent  to  the  path  of  the  particle.  Since  the  resolved  part  of  P 
in  any  direction  is  equal  to  the  algebraic  sum  of  the  resolved  parts  of 
Px,  P2,     .     .     .     in  that  direction, 

P  cos  0  =  Px  cos  6X  +  P,  cos  *,+    >.../'., 

Hence  (Pcos  d)ds  =  (Px  cos  dx)ds  +  (/>  cos  d2)ds  +     .     .     .     ; 
and 

C  (P  cos  d)ds  =  f   (Px  cos  6x)ds  -f  f   (/»,  cos  0.2)ds  +  .    .    .    . 

The  first  member  of  this  equation  represents  the  work  done  by  Pt 
while  the  successive  terms  of  the  second  member  represent  the  quan- 
tities of  work  done  by  Plf  P2i  .  .  .  ,  respectively  (Art.  350). 
Hence  the  proposition  is  proved. 

Examples. 

v  1.  A  body  weighing  50  lbs.  slides  a  distance  of  8  ft.  down  a  plane 
inclined  200  to  the  horizontal,  against  a  constant  retarding  force  of 
4  lbs.  due  to  friction.  Compute  the  total  work  done  upon  the  body 
by  its  weight  and  the  friction.  Ans.    104.8  foot-pounds. 

2.  Prove  that  the  total  work  done  by  gravity  upon  a  system  of 
particles  during  any  motion  has  the  same  value  as  if  the  entire  mass 
were  concentrated  at  the  center  of  gravity. 

Let  the  system  consist  of  particles  whose  weights  are  Wx ,  W2 , 

.     and  whose  elevations  above  a  certain  horizontal  plane  are 

initially  glt  st,     .     .     .      ;  and  let  their  elevations  above  the  same 

plane  in  the  final  positions  be  zx\  gJ,     ....     The  total  work  done 

by  gravity  upon  all  the  particles  is 

Wx(zx  -  O  +  W,{z2  —  gi)  +     .     .     . 

^(wxzx+w%z,+  .  .  o-(»ri^+;»w+  .  •  •). 

But  if  the  elevation  of  the  center  of  gravity  changes  from  2  to  z\ 
Wxzx+W2z2+     .     .     .     =(IVX+W2+     .     .     .     )2; 

Wxzx'  +  w2z.;  +    .    .    .    ==(&V+Wr,+    .    .    .    Yz'\ 

and  the  above  value  of  the  work  is  therefore  equal  to 

(j#V+     IV2+    .        .        .     )(2-2'), 

which  is  the  same  as  the  work  which  would  be  done  by  gravity  on 
the  whole  mass  if  concentrated  at  the  center  of  gravity.  Evidently 
the  same  proposition  holds  for  any  body  or  system  of  bodies. 


306  THEORETICAL    MECHANICS. 

353.  Total  Work  Done  on  a  Particle  Expressed  in  Terms  of 
Its  Initial  and  Final  Velocities  and  Its  Mass. —  Let  a  particle 
describe  any  path  under  the  action  of  any  forces.  Let  P  be  the 
value  of  the  resultant  of  all  the  forces  at  any  instant  and  6  the  angle 
between  P  and  the  tangent  to  the  path. 
Let  m  be  the  mass  of  the  particle  and  v  its 
velocity.  The  tangential  component  of  the 
acceleration  is  dv/dt,  and  the  tangential 
component  of  the  resultant  force  is  P  cos  6. 
Hence  there  may  be  written  the  equation 
of  motion  (Art.  289) 

FlG'  I57'  P  cos  0  ==  m(dv/dt). 

But  dvjdt  =  (dv/ds)(ds/dt)  ==  v{dv\ds)\  hence 

P  cos  6  =  tnv(dvlds),     or     (P  cos  0)ds  =  mv  dv. 

Let  s'  and  s"  be  the  values  of  s  for  any  two  positions  A  and  B, 
v'  and  v"  being  the  corresponding  values  of  v.     Then 

JCPcos  6)ds  =  f  tmtdv  =  \m(z>"2  —  v'2). 
t  *>  V' 

The  first  member  of  this  equation  is  the  total  work  done  on  the  par- 
ticle by  all  forces  during  the  motion  AB.  The  last  member  depends 
only  upon  the  values  of  the  velocity  of  the  particle  at  A  and  at  B. 

Vis  viva. — The  quantity  mv'1  is  often  called  the  vis  viva  of  the 
particle.  Using  this  term,  the  above  equation  expresses  the  propo- 
sition that  the  increase  in  the  vis  viva  of  a  particle  is  equal  to  twice 
the  total  work  done  upon  it.  This  is  often  called  the  principle  of 
vis  viva. 

Since  the  development  of  the  theory  of  energy,  the  term  vis  viva 
has  largely  given  place  to  another,  half  the  vis  viva  being  called  the 
"  kinetic  energy."  The  significance  of  this  name  will  be  explained 
below. 

The  above  principle  should  be  applied  in  the  solution  of  the  fol- 
lowing examples. 

Examples. 

v  1.  A  body  whose  mass  is  20  lbs.  falls  freely  under  the  action  of 
gravity.  How  far  does  it  fall  while  its  velocity  changes  from  40  ft.  - 
per-sec.  to  50  ft.-per-sec?  Ans.   About  14  ft. 

2.   A  body  is  projected  horizontally  with  a  velocity  of  100  ft. -per- 


WORK    AND    ENERGY.  307 

sec.     If  acted  upon  by  gravity  alone,  at  what  rate  will  it  be  moving 
when  20  ft.  lower  than  the  point  of  projection  ? 

A  us.    1 06. 2  ft.  -per-sec. 
3.   A  body  falls  vertically  from  a  height  of  5,000,000  ft.  above  the 
earth's  surface.    With  what  velocity  will  it  reach  the  earth,  neglecting 
the  resistance  of  the  air?     [Use  the  result  of  Ex.  2,  Art.  351.] 

A  n s.   1 6, 1 00  ft.  -per.  -sec. 


§  3.  Energy  of  a  Particle. 

354.  Meaning  of  Work  Done  by  a  Body. —  When  a  body  moves 
against  a  resisting  force,  it  is  said  to  do  work  against  that  force.  If  the 
point  of  application  of  the  force  is  displaced  in  a  direction  exactly 
opposite  to  that  of  the  force,  the  quantity  of  work  done  by  the  body 
is  equal  to  the  product  of  the  magnitude  of  the  force  into  the  dis- 
placement. If  the  displacement  has  some  other  direction,  the  effec- 
tive displacement  (or  projection  of  the  actual  displacement  on  a  line 
parallel  to  the  force)  must  be  used  in  computing  the  work. 

Thus,  a  body  thrown  vertically  upward  does  work  against  the 
force  of  gravity  while  rising,  the  quantity  of  work  being  equal  to  the 
product  of  the  weight  of  the  body  into  the  distance  it  ascends.  If  it 
moves  upward  along  an  inclined  plane,  the  effective  displacement  is 
the  projection  of  the  actual  displacement  upon  a  vertical  line.  A  body 
moving  through  the  air  in  any  direction  moves  against  the  resisting 
force  due  to  the  air,  and  thus  does  work  against  that  force.  A  body 
projected  along  a  horizontal  plane  moves  against  the  force  of  fric- 
tion ;  in  moving  any  distance  the  body  does  an  amount  of  work 
equal  to  the  product  of  the  frictional  force  into 
the  distance  the  body  moves.  & 

Let  P  be  the  magnitude  of  a  constant  force       :  \- f\ 

applied  to  a  body  at  a  points   (Fig.  158),        j        \        \ 

and  let  A  receive  the  displacement  AB,  mak-     B'        A  P 

ing  the  angle  6  with  the  direction  of  P.     The  Fig.  158. 

effective  displacement  of  A  against  the  force 

is  AB'  —  — AB  cos  6.      Hence  the  work  done  by  the  body  against 

the  force  is  -  P  X  AB  cos  0. 

This  value  is  positive  when  6  lies  between  900  and  1800.  The 
conception  of  work  done  by  a  body  against  a  force  is,  however,  to  be 
so  extended  as  to  include  negative  values ;  although  it  is  only  when 


308  THEORETICAL    MECHANICS. 

the  effective  displacement  and  the  force  have  opposite  directions  that 
the  force  can  properly  be  said  to  ' '  resist ' '  the  motion  of  the  body. 
A  body  moving-  downward  is  thus  regarded  as  doing  negative  work 
against  gravity  ;  in  algebraic  language,  the  force  of  gravity  is  a  neg- 
ative resistance. 

355.  Work  Done  by  a  Body  Always  the  Negative  of  Work 
Done  by  a  Force. —  From  the  foregoing  definitions  it  will  be  seen 
that  whenever  positive  work  is  done  by  a  body  against  a  force,  neg- 
ative work  is  done  by  the  force  upon  the  body ;  and  vice  versa.  The 
two  statements  are  merely  different  ways  of  expressing  the  same  fact. 
Whatever  displacement  the  point  of  application  of  a  force  may  re- 
ceive, the  work  may  be  computed  as  done  either  by  the  force  upon 
the  body  or  by  the  body  against  the  force.  The  two  quantities  are 
equal  in  magnitude  but  opposite  in  sign. 

All  the  results  above  found  in  the  discussion  of  the  work  done  by 
forces  upon  a  particle  may  be  extended  to  work  done  by  a  particle 
against  applied  forces,  by  merely  changing  the  sign  of  the  work  in 
every  case. 

356.  Energy  of  a  Body  Defined.—  When  the  condition  of  a 
body  is  such  that  it  can  do  work  against  a  force  or  forces  that  may 
be  applied  to  it,  the  body  is  said  to  possess  energy. 

The  quantity  of  energy  possessed  by  a  body  is  the  amount  of 
work  it  can  do  in  passing  from  its~  present  condition  to  some  standard 
condition. 

The  unit  of  energy  is  the  same  as  that  of  work. 

357.  Kinetic  Energy  of  a  Particle.—  If  a  particle  is  in  motion, 
it  will  continue  to  move  for  a  time,  however  great  a  resisting  force 
may  be  applied  to  it.  It  cannot  be  brought  to  rest  until  it  has  moved 
over  some  distance.  By  reason  of  being  in  motion,  therefore,  a  par- 
ticle can  do  work  against  any  force  or  forces  which  resist  the  motion  ; 
that  is,  it  possesses  energy.  Since  this  energy  is  possessed  by  the 
particle  by  reason  of  its  motion,  it  is  called  energy  of  motion  or 
kinetic  energy. 

A  single  particle  can  possess  no  energy  except  kinetic  energy. 
When  the  definition  of  energy  is  extended  to  systems  of  particles, 
another  form  of  energy  will  be  recognized. 

In  defining  the  quantity  of  energy  of  a  body  (Art.  356),  reference 
was  made  toa  "  standard  condition. ' '      The  full  meaning  of  ' '  con- 


WORK    AND    ENERGY.  309 

dition ' '  cannot  be  explained  until  the  theory  of  energy  is  extended 
to  systems  of  particles  and  bodies  in  general.  In  case  of  a  single 
particle,,  the  only  condition  affecting  its  capacity  to  do  work  is  its 
velocity  ;  the  ' '  standard  condition  "  is  a  standard  velocity.  So  long 
as  a  particle  possesses  any  velocity,  however  small,  it  can  move 
against  resisting  forces ;  in  estimating  the  energy  of  a  particle,  there- 
fore, the  standard  condition  will  be  taken  as  a  condition  of  rest. 

358.  To  Determine  the  Quantity  of  Kinetic  Energy  of  a  Par- 
ticle.—  It  will  now  be  shown  that  a  particle  of  given  mass  moving 
with  a  given  velocity  possesses  a  definite  quantity  of  kinetic  energy. 

In  order  to  determine  the  quantity  of  energy  possessed  by  a  par- 
ticle of  mass  tn  and  velocity  v,  it  is  necessary  to  determine  how  much 
work  will  be  done  by  the  particle  in  coming  to  rest.  Let  it  follow 
any  path  whatever  and  be  acted  upon  by  any  forces. 

By  the  principle  proved  in  Art.  353,  the  total  work  done  upon 
the  particle  while  its  velocity  changes  from  v'  to  v"  is 

\m{v"'l  —  v"\ 

The  total  work  done  by  the  particle  is  the  negative  of  this  quantity, 

In  coming  to  rest  from  a  velocity  v  the  particle  therefore  does  an 
amount  of  work  \mv2.     That  is, 

The  kinetic  energy  of  a  particle  is  equal  to  half  the  product  of  its 
mass  into  tJie  square  of  its  velocity. 

359.  Principle  of  Work  and  Energy  for  a  Particle. —  The  prin- 
ciple expressed  by  the  equations 

total  work  done  upon  particle  =  \rniv'"1  —  v'2)} 
total  work  done    by    particle  ==  \m(v'2 —  ^"2)> 

may  now  be  called  the  principle  of  ivork  and  energy  for  a  particle. 
Expressed  in  words, 

During  any  motion  of  a  particle  its  kinetic  energy  increases  by 
an  amount  equal  to  the  total  work  done  upon  the  particle  by  all 
forces ;  its  kinetic  energy  decreases  by  an  amount  equal  to  the  total 
work  done  by  the  particle  against  all  forces. 

These  two  statements  are  identical  in  meaning,  if  the  words  in- 
crease and  decrease  are  understood  in  their  algebraic  sense. 

This  principle  is  extremely  useful  in  solving  certain  problems. 


3IO  THEORETICAL    MECHANICS. 

The  equation  of  work  and  energy  can  often  be  written  immediately, 
the  work  being  expressed  in  terms  of  the  forces  acting  and  the  dis- 
tances through  which  they  act,  and  the  change  of  kinetic  energy  in 
terms  of  the  mass  of  the  particle  and  its  initial  and  final  velocities. 

It  must  be  remembered  that  a  kinetic  system  of  units  (Arts.  217, 
218)  must  be  employed  in  writing  the  equation,  since  it  is  deduced 
from  the  fundamental  equation  of  motion,  P  =  mp. 

Dimensions  of  kinetic  energy.  —  From  the  formula  mviJ2i  it  is 
evident  that  kinetic  energy  has  dimensions  ML2/T2.  This  agrees 
with  the  dimensions  of  work  when  kinetic  units  are  employed  (Art. 
346),  as  should  be  the  case,  since  the  equation 

work  =  change  of  kinetic  energy 

must  be  homogeneous. 

The  following  examples  should  be  solved  by  applying  the  prin- 
ciple of  work  and  energy.  The  value  of  g  (the  acceleration  due  to 
gravity  at  the  surface  of  the  earth)  is  taken  as  32.2  foot-second  units. 

Examples. 

1 .  Compute  the  kinetic  energy  of  a  body  of  20  lbs.  mass  moving 
at  the  rate  of  150  ft.-per-sec.  Ans.   6,990  foot-pounds. 

2.  If  the  direction  of  motion  of  the  body  in  Ex.  1  is  vertically 
upward  and  if  no  force  acts  upon  the  body  except  its  own  weight, 
how  high  will  it  rise  ?  (That  is,  how  far  must  it  move  against  gravity 
in  order  to  do  an  amount  of  work  equal  to  the  known  decrease  of 
kinetic  energy?) 

v  3.  If  the  resistance  of  the  air  is  constantly  equal  to  4  lbs.,  how 
high  will  the  body  rise?  Ans.  291  ft. 

v  4.  If  a  body  of  10  lbs.  mass  is  projected  horizontally  on  a  rough 
plane  with  a  velocity  of  50  ft.-per-sec,  how  far  will  it  move  before  its 
velocity  is  reduced  to  20  ft.  -per-sec. ,  the  retarding  force  due  to  fric- 
tion being  constantly  5  lbs.  ?  Ans.  65.2  ft. 

5.  A  body  weighing  10  lbs.  falls  vertically  under  gravity  against 
a  constant  force  of  1  lb.  due  to  the  resistance  of  the  air.  How  far 
must  it  fall  in  order  that  its  velocity  may  change  {a)  from  zero  to  20 
ft.-per-sec;  {b)  from  10  ft.-per-sec  to  20  ft.-per-sec.  ? 

Ans.  (a)  6.9  ft.     {b)  5.2  ft. 

6.  If  a  body  is  at  rest  at  a  distance  from  the  earth's  surface  equal 
to  the  radius  (taken  as  21,000,000  ft.),  and  falls  under  the  action  of 
its  own  weight,  what  will  be  its  velocity  on  reaching  the  surface  ? 

Ans.  26,000  ft.-per-sec 

7.  With  what  velocity  must  a  body  be  projected  from  the  surface 


WORK    AND    ENERGY.  3H 

of  the  earth  in  order  that  it  may  never  return,  no  force  except  the 
earth's  attraction  being  supposed  to  act?  Does  the  direction  of  pro- 
jection affect  this  result  ? 

Ans.  36,700  ft. -per-sec.  =  6.96  miles-per-sec. 

8.  A  particle  whose  mass  is  500  gr.  is  attached  to  one  end  of  an 
elastic  string  the  other  end  of  which  is  fixed.  The  ' '  natural  length  ' ' 
of  the  string  is  50  cm. ;  under  a  pull  equal  to  the  weight  of  500  gr. 
its  length  becomes  70  cm.  The  string  being  stretched  to  double  its 
natural  length,  the  particle  is  held  at  rest  and  then  released.  If  no 
force  acts  upon  the  particle  except  that  due  to  the  string,  what  great- 
est velocity  will  it  acquire?     (Take^  =  981  C.  G.  S.  units.) 

Ans.  350  cm. -per-sec. 

9.  Solve  Ex.  8,  taking  account  of  the  force  of  gravity,  the  string 
being  assumed  to  hang  vertically  downward. 

Ans.  210  cm. -per-sec 


§  4.  Energy  of  a  System. 

360.  Material  System  May  Possess  Energy  Not  Due  to  Mo- 
tion.—  The  definition  of  "energy  of  a  body  "  given  in  Art.  356  has 
thus  far  been  applied  only  to  a  single  particle.  When  the  definition 
is  applied  to  an  actual  body  or  to  an  aggregation  of  particles  re- 
garded as  a  system,  our  conception  of  energy  must  be  enlarged. 
Besides  the  kinetic  energy  which  may  be  possessed  by  the  individual 
particles,  the  system  as  a  whole  may  possess  energy  of  another  form. 

In  Part  III  will  be  developed  the  theory  of  the  motion  of  a  system 
of  particles,  including  the  general  theory  of  energy  for  any  material 
system.*  We  shall  here  give  an  elementary  explanation  of  the 
meaning  of  energy  of  a  system,  depending  only  upon  principles  de- 
veloped in  the  foregoing  chapters. 

361.  Energy  of  a  Deformed  Elastic  Body. —  An  example  of  a 
body  which  can  do  work  even  if  initially  at  rest  is  furnished  by  an 
elastic  string.  Such  a  string,  if  not  acted  upon  by  external  forces, 
assumes  a  certain  ' '  natural ' '  length.  By  the  application  of  equal 
and  opposite  forces  at  the  ends  it  may  be  stretched,  the  amount  of 
elongation  depending  upon  the  magnitude  of  the  applied  forces.  If 
the  forces  are  diminished,  it  shortens,  and  unless  the  stretching  has 
been  too  great  it  nearly  or  quite  resumes  its  original  ■ '  natural ' ' 
length  when  the  forces  cease  to  act. 

*  See  Chapter  XXIII. 


312  THEORETICAL    MECHANICS. 

While  in  the  stretched  condition,  the  string  possesses  energy  in 
accordance  with  the  definition  (Art.  356)  ;  its  condition  is  such  that 
it  can  do  work  against  applied  forces.  It  will,  in  fact,  do  work 
against  the  forces  applied  to  the  ends  if  these  are  gradually  dimin- 
ished so  as  to  permit  the  string  to  shorten. 

This  energy  is  not  kinetic ;  it  is  not  due  to  the  motion  of  the 
body  as  a  whole  nor  of  its  parts.  It  is  due  to  the  relations  of  the 
parts  of  the  body  to  one  another.  Just  what  these  relations  are  we 
do  not  know,  since  the  ultimate  structure  of  the  body  is  unknown. 
It  can  only  be  said  that  the  particles  tend  to  assume  certain  relative 
positions,  that  any  departure  from  these  positions  calls  into  action 
resisting  forces,  and  that  these  forces  will,  if  the  external  forces 
cease  to  act  or  decrease  in  magnitude,  bring  the  particles  back  to- 
ward these  positions. 

Any  elastic  body,  while  deformed  and  tending  to  resume  its 
1 '  natural ' '  shape,  possesses  energy  irrespective  of  its  condition  of 
motion.  Such  energy  may  be  called  "energy  of  position,"  since 
it  is  due  to  the  tendency  of  the  particles  of  the  body  to  assume  def- 
inite relative  positions. 

362.  Energy  Due  to  Gravity. —  A  body  near  the  earth  whose 
position  is  such  that  it  can  move  to  a  lower  level  can  do  work  against 
forces  resisting  its  descent.  The  condition  of  the  body  is  therefore 
such  as  to  satisfy  the  definition  of  energy  (Art.  356).  This  energy 
is,  however,  possessed  by  the  system  consisting  of  the  body  and  the 
earth  rather  than  by  the  body  alone.  It  is  the  weight  of  the  body 
that  enables  it  to  move  against  the  forces  which  resist  its  downward 
motion.  The  amount  of  work  which  can  be  done  against  the  resist- 
ing forces  is  just  equal  to  the  amount  of  work  done  on  the  body  by 
the  earth's  attraction  during  the  descent. 

This  energy,  like  that  of  a  deformed  elastic  body,  may  be  called 
energy  of  position,  since  it  depends  upon  the  relative  position  of  the 
bodies  constituting  the  system  and  upon  the  action  of  internal  forces 
tending  to  cause  a  definite  change  in  their  relative  position. 

Potential  energy  is  the  name  usually  given  to  energy  of  position. 

After  the  foregoing  illustrations  of  the  meaning  of  potential  en- 
ergy, it  will  be  well  to  restate  the  definition  of  energy  with  explicit 
reference  to  a  system  of  bodies  or  of  particles. 

363.  Energy  of  a  System  Defined. — A  system  of  bodies  or  of 
particles  is  said  to  possess  energy  when  its  condition  is  such  that 


WORK    AND    ENERGY.  313 

it  can   do   work   against  external  forces   which  may  be  applied 
to  it. 

The  quantity  of  energy  possessed  by  a  system  is  the  amount  of 
work  it  will  do  against  external  forces  in  passing  from  its  present 
condition  to  some  standard  condition. 

It  is  important  to  notice  the  word  "external"  in  this  definition. 
The  distinction  between  external  and  internal  forces  has  been  ex- 
plained in  Art.  119. 

What  shall  constitute  a  ' '  system ' '  of  bodies  or  of  particles  is  a 
matter  of  arbitrary  selection.  Thus,  we  may  regard  a  single  body  as 
forming  a  system,  or  we  way  regard  the  body  and  the  earth  as  form- 
ing a  system.  The  body  itself  possesses  kinetic  energy  if  in  motion  ; 
the  body  and  the  earth  regarded  as  a  system  possess  energy  because 
of  their  relative  position. 

Similarly,  the  particles  of  a  deformed  elastic  body  individually 
possess  no  energy  unless  they  are  in  motion  ;  collectively  they  pos- 
sess energy  because  of  the  internal  forces  which  tend  to  cause  the 
body  to  assume  a  definite  shape. 

These  cases  illustrate  the  general  principle  that  potential  energy 
is  due  to  the  action  of  internal  forces. 

364.  Choice  of  "  Standard  Condition." — It  is  now  clear  that  the 
words  "standard  condition"  in  the  definition  of  energy  refer  to  some 
definite  set  of  velocities  and  positions  of  the  members  of  the  system. 
The  actual  value  of  the  energy  depends  upon  what  is  chosen  as  the 
standard  condition.  This  choice  is  governed  solely  by  convenience. 
In  estimating  kinetic  energy  it  is  convenient  to  assume  rest  as  the 
standard  condition  for  each  particle,  since  the  algebraic  expression 
for  the  energy  is  thereby  simplified.  In  computing  potential  energy 
it  is  often  a  matter  of  indifference  what  set  of  positions  constitutes 
the  standard  condition.  In  particular  applications  we  are  concerned 
with  the  difference  between  the  values  of  the  energy  in  different  con- 
ditions rather  than  with  absolute  values ;  this  difference  is  the  same 
whatever  be  assumed  as  the  standard  condition. 

365.  Value  of  Potential  Energy  Due  to  Gravity. —  Potential 
energy  due  to  the  weights  of  bodies  near  the  earth  is  usually  briefly 
referred  to  as  energy  possessed  by  the  bodies  themselves  ;  although 
as  above  stated  it  is  strictly  possessed  by  the  system  of  which  the 
earth  and  the  bodies  are  members.     In  computing  its  value,  a  hori- 


314  THEORETICAL    MECHANICS. 

zontal  surface  or  ' '  datum  plane ' '  must  be  chosen  as  specifying  the 
"standard"  positions  of  the  bodies. 

It  may  be  shown  that  the  ' '  gravity ' '  potential  energy  possessed 
by  a  particle  is  equal  to  the  product  of  its  weight  into  its  height 
above  the  assumed  reference  plane ;  and  that  the  total  potential  en- 
ergy of  a  system  of  particles  is  the  same  as  if  the  entire  mass  were 
concentrated  at  the  center  of  gravity. 

Let  W  be  the  weight  of  a  particle  whose  height  above  datum  is 
Z.  Its  potential  energy  is  equal  to  the  work  it  can  do  against  resist- 
ing forces  in  descending  to  the  standard  position,  —  i.  e.,  to  the 
datum  plane.  If  the  particle  is  at  rest  in  the  initial  position  and 
comes  to  rest  in  the  standard  position,  the  work  it  has  done  against 
all  forces  except  its  weight  is  equal  to  the  work  done  upon  it  by 
its  weight  (Art.  359).  The  value  of  this  work  is  Wz ;  which  is  there- 
fore the  value  of  the  potential  energy  of  the  particle  (strictly,  of  the 
earth  and  the  particle  regarded  as  a  system)  when  it  is  at  a  height  z 
above  its  standard  position. 

Again,  let  a  system  consist  of  particles  whose  weights  are  W\ , 
W2 ,  .  .  .  ,  at  heights  zx ,  z% ,  \  .  .  above  the  plane  of  ref- 
erence, and  let  z  be  the  height  of  the  center  of  gravity  of  the  system 
above  the  same  plane.     Then 

( w,  +  m  +   .   .  .   )*  =  wxzx  +  w,z.2  +   .   .   .   . 

The  second  member  of  this  equation  is  the  total  potential  energy  of 
all  the  particles  in  the  supposed  positions,  and  the  first  member  is 
the  potential  energy  of  the  system  on  the  assumption  that  all  the  par- 
ticles are  concentrated  at  the  center  of  gravity. 

Potential  energy  due  to  the  weights  of  bodies  and  to  their  posi- 
tions on  the  earth  is  one  of  the  most  important  forms  of  energy  with 
which  we  are  concerned  practically.  In  the  case  of  streams  of  water, 
this  kind  of  energy  is  available  for  the  performance  of  useful  work. 
The  amount  of  available  energy  depends  upon  two  factors,  —  weight 
of  water  and  available  fall. 

366.  Principle  of  Work  and  Energy  for  a  Material  System. — 

During  any  change  of  condition  of  a  material  system,  its  total  en- 
ergy (kinetic  and  potential}  increases  by  an  amount  equal  to  the  total 
work  done  upon  the  members  of  the  system  by  external  forces ;  its 
total  energy  deceases  by  an  amount  equal  to  the  total  work  done  by 
its  members  against  external  forces. 


WORK    AND    ENERGY.  315 

This  principle  is  in  accordance  with  the  definition  of  energy  of  a 
system  (Art.  363).  Thus,  consider  the  system  consisting  of  the 
earth  and  a  particle  whose  mass  is  M  pounds.  If  the  height  of  the 
particle  above  its  standard  position  is  z  feet  and  if  its  velocity  is  v  feet 
per  second,  the  total  energy  of  the  system  is 

Mz  -f-  Mv2J2g  foot-pounds. 

Suppose  the  elevation  and  velocity  of  the  particle  change,  their  initial 
values  being  glt  vx,  and  their  final  values  z2 ,  vt.  If  no  external  force 
(/'.  e. ,  no  force  except  its  weight)  acts  upon  the  particle  during  the 
change,  the  principle  of  Art.  359  shows  that 

Aft*  -  Bt)  =  mpf  -  v*)lig% 

or  M{zy  +  vlJ2g)  =  M(z,  +  vflTg)  ; 

that  is,  the  final  value  of  the  total  energy  of  the  system  is  equal  to 
its  initial  value.  But  if  the  motion  is  resisted  by  an  external  force 
against  which  the  particle  does  W  foot-pounds  of  work,  the  principle 
of  Art.  359  gives  the  equation 

M(Si  -  *0  -  W  =  M(v*  -  v^\2g, 

or  M{zx  +  V*l2g)  —  W  ==  MQtt  +  vil2g). 

That  is,  the  system  has  lost  W  foot-pounds  of  energy. 

As  another  illustration,  consider  a  deformed  elastic  body.  Let 
its  ' '  natural ' '  shape  be  taken  as  the  standard  condition  for  estimating 
its  potential  energy,  and  suppose  it  to  be  so  deformed  that  in  return- 
ing to  the  natural  shape  it  can  do  E  units  of  work  against  external 
forces  ;  then  (assuming  its  particles  to  possess  no  energy  of  motion) 
E  is  its  total  energy.  Now  suppose  the  deformation  to  decrease 
gradually  until  the  body  has  done  W  units  of  work  against  the  ex- 
ternal forces  ;  it  can  still  do  E  —  W  units  of  work  in  coming  to  the 
standard  condition,  so  that  its  energy  is  now  E  —  IV.  That  is,  its 
energy  has  decreased  by  an  amount  equal  to  the  work  it  has  done 
against  external  forces.  * 

*  The  reader  may  notice  that  the  principle  of  work  and  energy  for  a  mate- 
rial system, —  in  fact  the  very  supposition  that  potential  energy  has  any  definite 
value, — involves  the  assumption  that  the  total  work  done  by  the  internal 
forces  during  any  change  of  condition  has  a  definite  value  which  is  independ- 
ent of  the  way  in  which  the  change  of  condition  takes  place.  For  a  rigorous 
discussion  of  this  assumption,  see  Chapter  XXIII. 


3l6  THEORETICAL    MECHANICS. 

367.  Machine. —  A  machine  has  been  defined  (Art.  in)  as  a 
device  for  the  application  of  force.  In  most  cases,  however,  the  pro- 
duction of  motion  is  a  part  of  the  object  of  a  machine.  Stated  com- 
pletely, the  function  of  a  machine  is  to  do  work  against  external 
forces. 

Thus,  a  system  of  pulleys  used  to  lift  a  heavy  body  against  gravity 
does  work  against  the  force  which  the  body  exerts  upon  the  system 
by  reason  of  its  weight. 

A  steam-engine  is  designed  to  cause  certain  bodies  connected 
with  it  to  move  against  resisting  forces  ;  in  accomplishing  this  object 
the  moving  parts  of  the  engine  do  work  against  the  forces  which 
these  bodies  exert  upon  them. 

In  doing  work  a  machine  continually  gives  up  energy  (Art.  366), 
and  it  cannot  continue  to  operate  unless  energy  is  supplied  to  it.  The 
work  done  upon  it  by  one  set  of  external  forces  must  on  the  whole 
be  equal  to  the  work  done  by  it  against  another  set.  Thus,  the  ex- 
ternal force  which  is  doing  work  upon  a  steam-engine  is  the  pressure 
of  the  steam  upon  the  piston.  The  external  forces  against  which  the 
engine  is  doing  work  are  {a)  the  tension  in  the  belt  which  connects 
it  with  the  driven  machinery,  and  (J?)  the  frictional  resistances  to  the 
motion  of  the  parts  of  the  engine.  The  external  work  done  by  the 
engine  is  thus  only  in  part  utilized,  that  done  against  frictional  resist- 
ances being  lost.* 

Efficiency.  —  The  efficiency  of  a  machine  is  the  ratio  of  the  useful 
work  done  by  it  to  the  energy  supplied  to  it.  The  efficiency  is  always 
less  than  unity. 

368.  Power  or  Activity.— The  rate  at  which  a  machine  does 
work  is  called  its  power  or  activity. 

It  would  be  natural  to  take  as  the  unit  of  power  that  correspond- 
ing to  a  unit  of  work  done  per  unit  time  ;  as  a  foot-pound  per  second, 
or  a  meter-kilogram  per  second,  or  an  erg  per  second.  Such  a  unit 
is  too  small  to  be  convenient  for  ordinary  use. 

The  ordinary  British  unit  is  the  horse-power,  equal  to  550  foot- 
pounds per  second. 

The  French  horse-power,  or  force  de  cheval,  is  equal  to  75 
meter-kilograms  per  second. 

*The  energy  represented  by  the  work  done  against  friction  is  "lost" 
only  in  the  sense  of  being  unavailable  for  any  useful  purpose.  In  reality  this 
energy  is  transformed, — principally  into  heat-energy.     See  Chapter  XXIII. 


WORK    AND    ENERGY.  317 

The  unit  of  activity  commonly  employed  in  electrical  engineering 
is  the  watt,  or  the  kilozvatt.  A  watt  is  defined  as  io7  ergs  per  sec- 
ond, a  kilowatt  being  therefore  io10  ergs  per  second. 

Examples. 

v  1.  A  stream  of  water  is  20  ft.  wide,  its  average  depth  is  3  ft.,  and 
the  average  velocity  in  the  cross-section  is  3  miles  per  hour.  If  there 
is  an  available  fall  of  200  ft. ,  how  much  available  potential  energy  is 
possessed  by  one  minute' s  supply  of  water  (or  strictly,  by  the  system 
consisting  of  the  water  and  the  earth)  ?  Assume  the  density  of  water 
to  be  62.5  lbs.  per  cu.  ft. 

Ans.  One  minute's  supply  of  water  is  15,840  cu.  ft.  or  990,000 
lbs.  In  falling  200  ft.  this  water  gives  up  198,000,000  foot-pounds 
of  potential  energy. 

2.  In  Ex.  1,  compute  the  kinetic  energy  of  one  minute's  supply 
of  water.  Ans.   297,600  foot-pounds. 

3.  What  available  H.  P.  is  represented  by  the  stream  described  in 
Ex.  1?  An s.  6,000  H.  P.,  neglecting  kinetic  energy. 

4.  Water  is  to  be  lifted  150  ft.  at  the  rate  of  5  cu. -ft. -per-sec. 
What  effective  H.  P.  must  be  realized  by  the  engine  and  pump  ? 

Ans.  85.2  H.  P. 
V       5.  A  nozzle  discharges  a  stream  1  in.  in  diameter  with  a  velocity 
of  80  ft. -per-sec.     (a)  How  much  kinetic  energy  is  possessed  by  the 
amount  of  water  which  flows  out  in  1  min.?     (fr)  If  this  energy  could 
all  be  utilized  by  a  water-wheel,  what  would  be  its  power  ? 

Ans.  (a)  162,800  foot-pounds.     (£)  4.93  H.'P. 
\       6.   In  Ex.  5,  suppose  the  jet  to  drive  a  water-wheel  connected  with 
a  pump  which  lifts  water  20  ft.     If  the  efficiency  of  the  whole  appara- 
tus is  0.48,  how  much  water  is  lifted  per  minute? 

Ans.  62.5  cu.  ft. 

7.  A  well  4  ft.  square  and  85  ft.  deep  is  excavated  in  earth  weigh- 
ing 180  lbs.-per-cu.-ft.  How  much  work  is  done  in  lifting  the  earth 
to  the  surface  ? 

8.  Show  that  the  relative  values  of  the  units  of  activity  above  de- 
lined  are  as  follows  : 

1  horse-power  =  746  watts  =  1. 01 385  force  de  cheval. 
1  kilowatt  =1.34  horse-power  =1.36  force  de  cheval. 

9.  In  Ex.  5,  let  the  energy  of  the  jet  be  employed  in  producing 
an  electric  current  by  means  of  a  water-wheel  driving  a  generator. 
If  the  efficiency  of  the  water-wheel  is  0.75  and  that  of  the  generator 
0,85,  what  power  in  kilowatts  is  represented  by  the  current? 


318  THEORETICAL    MECHANICS. 

§  5.   Virtual  Work. 

369.  Work-Condition  of  Equilibrium. —  Let  a  particle,  acted 
upon  by  any  number  of  forces,  receive  any  displacement.  The  work 
done  by  the  resultant  of  the  system  is  equal  to  the  algebraic  sum  of 
the  quantities  of  work  done  by  the  several  forces.  (Art.  352.)  If 
the  forces  are  in  equilibrium,  their  resultant  is  zero  and  the  work 
done  by  it  is  zero.     Therefore, 

If  a  system,  of  con  air  rent  forces  is  in  equilibrium,  the  algebraic 
sum  of  the  quantities  of  work  done  by  them  during  any  displacement 
of  their  point  of  application  is  zero. 

370.  Virtual  Displacement  of  a  Particle. —  Whatever  forces 
may  be  acting  upon  a  particle,  it  may  at  any  instant  be  at  rest,  or  it 
may  be  moving  in  any  direction.  Its  displacement  during  a  short 
interval  may  therefore  have  any  direction  ;  its  actual  direction  being 
determined  by  previous  conditions.  The  principle  stated  in  the  last 
Article  is  not  restricted  to  the  actual  displacement,  but  is  true  for  any 
arbitrary  hypothetical  displacement. 

Any  hypothetical  displacement  of  a  particle  is  called  a  virtual 
displacement.  It  may  or  may  not  coincide  with  the  actual  displace- 
ment. 

371.  Principle  of  Virtual  Work.— The  work  done  by  a  force 
during  a  supposed  (or  virtual)  displacement  of  its  point  of  application 
is  called  its  virtual  work.  The  principle  of  Art.  369  may  therefore 
be  stated  as  follows  : 

If  a  system  of  concurrent  forces  is  in  equilibrium,  tJie  algebraic 
sum  of  their  virtual  works  is  zero  for  any  possible  displacement. 

In  applying  this  principle,  it  is  common  to  assume  the  displace- 
ment to  be  infinitesimal.  This  is  not  necessary,  however,  if  the  forces 
are  assumed  to  remain  constant  in  magnitude  and  direction  through- 
out the  displacement. 

372.  Conditions  of  Equilibrium.  — The  conditions  of  equilibrium 
for  concurrent  forces  may  be  deduced  from  the  principle  of  virtual 
work. 

Let  A  (Fig.  1 59)  be  the  point  of  application  of  several  forces  in 
equilibrium,  and  let  it  receive  a  displacement  AB  =  //.  The  work 
done  by  a  force  of  magnitude  P,  acting  in  a  direction  inclined  at 
angle  6  to  AB,  is  Ph  cos  6.     For  any  number  of  forces,  Plt  P%t 


WORK    AND    ENERGY.  319 

.     .      .      ,  acting   at    angles  0X,    6.n      .      .      .     with   AB,    the  total 

work  is 

(Px  cos  0,  +  Pt  cos  0t  +     :     .     .     )/*. 

If  the  forces  are  in  equilibrium,  this  work  is  zero  ;  therefore 
Px  cos  0,  +  P.,  cos  $%  +     .     .     .     =  o. 

This  is  identical  with  the  equation  obtained  by  resolving  forces  par- 
allel to  AB. 

By  taking  virtual  displacements  in  different  directions,  any  number 
of  equations  of  the  above  form  may  be  obtained, 
just  as  by  resolving  forces  in  different  directions. 

It  is  evident  that,  for  coplanar  forces,  only 
two  of  these  possible  equations  can  be  independ- 
ent. For  if  the  total  virtual  work  is  zero  for 
any  displacement,  the  only  possible  direction  for 
the  resultant  force  is  perpendicular  to  that  dis- 
placement ;  and  if  the  total  virtual  work  is  zero 
for  each  of  two  displacements  which  are  not  parallel,  the  resultant 
must  be  zero. 

373.  Forces  in  Three  Dimensions. —  The  principle  of  virtual 
work  is  readily  seen  to  hold  for  forces  in  three  dimensions.  For  the 
proposition  upon  which  it  depends,  —  that  the  work  done  by  the 
resultant  of  any  forces  is  equal  to  the  sum  of  the  quantities  of  work 
done  by  the  several  forces, —  is  true  without  restricting  the  forces  to 
a  plane.  This  is  evident  from  Art.  352.  It  may  also  be  seen  by 
applying  the  principle  first  to  two  forces  and  their  resultant  (which 
are  coplanar),  then  to  this  resultant  and  a  third  force,  and  so  on. 

374.  Application  of  Principle  of  Virtual  Work. —  Any  prob- 
lem in  equilibrium  of  concurrent  forces  may  be  solved  by  the  principle 
of  virtual  work.  The  method  has  no  advantage  over  the  usual 
method  of  resolving  forces,  so  long  as  only  a  single  system  of  con 
current  forces  is  involved.  It  is,  however,  useful  in  the  solution  of 
many  problems  relating  to  systems  of  particles,  and  systems  of  rigid 
bodies.     (See  Chapter  XXIII.) 

In  applying  the  principle  of  virtual  work  to  a  particle,  any  force 
may  be  eliminated  from  the  equation  of  work  by  choosing  the  dis- 
placement in  such  direction  that  the  virtual  work  of  that  force  is 
zero  ;  that  is,  at  right  angles  to  the  direction  of  the  force. 


320 


THEORETICAL   MECHANICS. 


Examples. 

t.  A  particle  rests  on  a  smooth  inclined  plane  under  the  action  of 
a  horizontal  force.      Required  the  relation  between  the  supporting 

force  and  the  weight  of  the  particle. 

Let  a  =  inclination  of  plane  to  hori- 
zontal, W  =  weight  of  body,  P  =  hori- 
zontal force,  N  =  normal  reaction  of  plane 
(Fig.  1 60). 

In  order  to  eliminate  the  unknown  force 
Ny  assume  a  displacement  along  the  plane. 
Let  its  direction  be  up  the  plane  and  its 
length  AB  =  h.  The  virtual  displacement 
of  P  is  +  h  cos  a  ;  that  of  W  is  — h  sin  a  ; 
and  the  equation  of  virtual  work  is 

Ph  cos  a  —   Wh  sin  a  =  o, 
from  which  P  =  W  tan  a. 

The  value  of  N  may  be  found  by  taking  a  displacement  at  right 
angles  to  P. 

2.  A  bead  whose  weight  is  W  is  free  to  slide  on  a  smooth  circu- 
lar wire  in  a  vertical  plane.  A  string  attached  to  the  bead  passes 
over  a  smooth  peg  at  the  highest  point  of  the  circle  and  sustains  a 
weight  P.     Determine  the  position  of  equilibrium. 

In  Fig.  161,  AB  represents  the  vertical  diameter  of  the  circle,  C 
the  bead,  O  the  center  of  the  circle.     Let  0 
=  angle  CA  O,  20  =  angle  COB.    The  bead  A 

is  acted  upon  by  three  forces  :  W  vertically 
downward,  P  in  direction  CA,  normal  pres- 
sure N  in  direction  CO.  Take  a  displace- 
ment h  upward  along  the  circle.  The  equa- 
tion of  virtual  work  is 

Ph  sin  6  —Wh  sin  2d  =  o. 
Solving,  cos  0  =  P/2  W. 

This  and  the  preceding  problem  both  ex- 
emplify the  statement  above  made,  that  the 
equation  of  virtual  work  possesses  no  advan- 
tage over  the  equation  obtained  by  resolving 
forces.      After  canceling   the  displacement,  Fig.  161. 

which  is  a  common  factor  in  every  term  of 

the  work  equation,  this  becomes  identical  with  the  equation  obtained 
by  resolving  forces  in  the  direction  of  the  displacement. 


Part  III. 

MOTION    OF    SYSTEMS    OF    PARTICLES   AND 
OF   RIGID   BODIES. 


CHAPTER    XVIII. 

MOTION   OF    ANY    SYSTEM    OF    PARTICLES. 

§  i .   Motions  of  Individual  Particles  of  a  System. 

375.  Material  System  Defined. —  In  Part  II  we  have  been  con- 
cerned mainly  with  the  motion  of  a  single  particle.  An  important 
part  of  the  discussion  has  related  to  the  problem  of  determining  the 
motion  of  a  particle  when  the  forces  applied  to  it  are  known.  In  the 
definition  of  force  (Arts.  32  and  212)  the  fact  was  emphasized  that  a 
force  is  an  action  exerted  by  one  particle  upon  another ;  so  that  the 
dynamical  equation  (force  =  mass  X  acceleration)  for  any  particle 
always  implies  that  the  motion  of  that  particle  is  influenced  by  other 
particles.  But  the  attention  has  usually  been  directed  particularly 
to  a  single  particle. 

It  is  often  desirable  to  consider  the  motions  of  several  particles 
collectively  instead  of  individually.  Any  number  of  particles  thus 
treated  as  a  group  may  be  called  a  material  system. 

It  is  found  that  by  applying  the  fundamental  laws  of  motion 
already  explained,  certain  important  general  principles  can  be  de- 
duced relating  to  the  motion  of  any  system  of  particles. 

The  following  analysis  will  be  restricted  mainly  to  the  case  in 
which  the  motion  is  confined  to  a  fixed  plane.  It  will  be  pointed 
out  at  the  end  of  the  chapter  that  most  of  the  principles  deduced  are 
true  without  this  restriction. 

376.  Coordinates  of  Position. — Two  coordinates  suffice  to  specify 
the  position  of  any  particle  in  a  given  plane.  Rectangular  coordi- 
nates will  be  employed  in  the  following  discussion. 

Let  the  system  consist  of  any  number  of  particles,  their  masses 
being  ;«1(  m.2i     .     .     .     ,  and  their  coordinates  of  position  {xxyy^)t 

21 


322  THEORETICAL    MECHANICS. 

C^2,j^2),  ....  The  motion  is  completely  known  if  the  coor- 
dinates of  every  particle  are  known  functions  of  the  time. 

377.  Motion  of  Mass-Center.— The  center  of  mass  plays  an 
important  part  in  the  theory  of  the  motion  of  a  system  of  particles. 
If  its  coordinates  are  (x,  y),  its  position  is  given  by  the  equations 

(?nx  +  m2-{-  .  .  .  )x  =  mxxx  +  m2x2  +  •  •  •  ,  .  (1) 
(mx  -f  m2 +     .     .     .     )y  =  mxyx  +  m2y2  +    .    .    .    ,  .     (2) 

as  in  Art.  152. 

If  the  coordinates  of  every  particle  are  known  functions  of  the 
time,  x  and  y  are  also  known  functions  of  the  time  ;  that  is,  the  mo- 
tion of  the  mass-center  is  known  from  equations  (1)  and  (2)  as  soon 
as  the  motion  of  every  particle  is  known. 

By  differentiating  these  equations,  the  velocity  of  the  mass-center 
may  be  determined.  Its  x-  and  j-components  are  dxjdt,  dyjdt,  given 
by  the  equations 

{mx  +  m2  -f-  .  .  .  )(dx/dt)  =  mxXx  +  m2x2  +  .  .  .  ,  (3) 
{mx  +  m2  +  •  •  •  )(dyldf)  =  mxyx  -f-  m^y,  -f  ....  (4) 

The  axial  components  of  the  acceleration  of  the  mass-center  are 
d2xldt2y  d2y/dt'\  given  by  the  equations 

(m,  -f  m2  +    .     .     .    )(d'2x/dt2)=  mxxx+  m2x2+    .     .     .     ,     (5) 

{mx  +  *•,+_.;.■.    )(d2y/dt2)  =  mxyx  -f  ;;/2>2  +   ....     (6) 

378.  Equations  of  Motion  for  Individual  Particles. —  If /^de- 
notes the  resultant  of  all  the  forces  acting  upon  a  particle  of  mass  mf 
and/>  the  resultant  acceleration,  the  general  equation  of  motion  for 
the  particle  (Art.  256)  is 

P  =  mp. 

Such  an  equation  may  be  written  for  every  particle  of  a  system.  Let 
Px  denote  the  resultant  of  all  forces  acting  upon  the  particle  whose 
mass  is  mx ,  and  px  its  resultant  acceleration  ;  let  P2  denote  the  re- 
sultant force  and  p2  the  resultant  acceleration  for  the  particle  of  mass 
m2 ;  with  similar  notation  for  every  particle  of  the  system  ;  then  there 
may  be  written  the  equations 

Px  =  mxpx,     P2  =  m2p2,     ....  .     (1) 

Each  of  these  is  a  vector  equation,  expressing  identity  of  direction  as 
well  as  equality  of  magnitude, 


MOTION   OF    ANY   SYSTEM    OF    PARTICLES.  323 

If  the  motion  is  restricted  to  a  plane,  each  of  equations  ( 1 )  yields 
two  independent  equations,  obtained  by  resolving  in  two  directions. 
If  these  directions  are  those  of  the  rectangular  axes  of  x  and  y,  the 
equations  may  be  written  as  follows  : 

Xx  =  my'xx,      Vi  =  wiil'i 

X,  =  mtxtl      Y,  =  m,y2 ;  >  (2) 


In  these  equations  Xx  and  Y1  are  the  axial  components  of  Px ;  xx 
and  yx  the  axial  components  oipx ;  etc. 

The  complete  determination  of  the  motion  of  every  particle  re- 
quires the  solution  of  this  system  of  simultaneous  differential  equa- 
tions. In  order  that  this  may  be  possible,  the  forces  acting  upon 
every  particle  must  be  known  functions  of  the  coordinates  and  the 
time. 

The  forces  acting  upon  any  particle  are  in  general  part  external 
(exerted  by  particles  not  belonging  to  the  system)  and  part  internal 
(exerted  by  other  particles  belonging  to  the  system).  The  values 
of  Px,  P2,  .  .  .  and  of  their  axial  components  include  all 
forces,  both  external  and  internal.  In  many  cases  it  is  possible  to 
express  the  values  of  these  forces  in  terms  of  the  coordinates.  But 
only  in  exceptional  cases  are  the  resulting  equations  sufficiently 
simple  to  admit  of  complete  integration.  It  is  possible,  however,  by 
properly  combining  equations  (2),  to  derive  certain  general  equa- 
tions which  throw  light  upon  the  motion  of  the  system  as  a  whole. 

379.  Elimination  of  Internal  Forces. —  If  the  equations  obtained 
by  resolving  along  the  ^r-axis  be  added,  the  result  is 

Xx+X%  +     .     .     .    =  mlx1  +  mtxs  + (3) 

Similarly,  from  the /-equations, 

vi  +  Yt  -[-...=  mjx  +  m2y2  + (4) 

The  first  members  of  these  equations  include  the  axial  components 
of  all  forces,  external  and  internal,  acting  upon  every  particle  of  the 
system. 

By  Newton's  third  law,  the  two  forces  which  any  two  particles 
exert  upon  each  other  are  equal  and  opposite  ;  their  components  in 
any  direction  are  therefore  equal  and  opposite,  and  the  sum  of  these 
components  is  zero.     In  the  sum 


324  THEORETICAL    MECHANICS. 

x,  +  x,  +   .   .   . 

the  .r-components  of  the  internal  forces  may  therefore  be  omitted, 
their  sum  for  the  whole  system  being  zero.  The  same  is  true  of  the 
/-components.  Therefore,  if  X  and  Fare  the  axial  components  of 
the  vector  sum  of  the  external  forces  applied  to  all  particles  of  the 
system,  equations  (3)  and  (4)  may  be  written 

X  =  m1xl  -f  f**x%  +'*••• 


(5) 
Y  =  mxyx  -f  mtyt  -f 

380.  Equations  of  Motion  of  Mass-Center. — Using  the  values 
of  the  axial  components  of  the  acceleration  of  the  center  of  mass, 
given  in  Art.  377,  the  above  equations  may  be  written 

X=(ml  +  ma+     .     .     .     )(d2x/dt'2) 


Y=(m1  +  m.2^      .     .     .     ){d*j/A*).m 

Equations  (6)  show  that  the  acceleration  of  the  mass-center  of 
the  system  is  equal  to  that  of  a  particle  of  mass  (ml  +  m2  -f-  .  .  .  ) 
acted  upon  by  forces  whose  resultant  has  axial  components  X  and  Y. 
In  other  words, 

If  a  particle  of  mass  equal  to  the  total  mass  of  the  system  were 
acted  upon  by  forces  equal  and  parallel  to  the  external  forces  applied 
to  the  system,  its  acceleration  would  be  equal  to  the  actual  accelera- 
tion of  the  mass-center  of  the  system. 

In  order  to  determine  the  motion  of  the  center  of  mass,  it  is 
therefore  needful  only  to  know  the  external  forces. 

Examples. 

1.  Two  particles,  of  masses  50  lbs.  and  40  lbs.,  are  acted  upon  at 
a  certain  instant  by  parallel  forces  of  75  poundals  and  60  poundals 
respectively,  whose  lines  of  action  are  4  ft.  apart  and  perpendicular 
to  the  line  joining  the  particles.  Determine  (a)  the  position  of  the 
mass-center  and  (b)  its  acceleration  at  the  instant  named. 

Ans.  (b)  1.5  ft.-per-sec.-per-sec,  if  the  forces  have  the  same 
direction. 

2.  If  the  two  particles  of  Ex.  1  attract  each  other  with  forces  of 
40  poundals,  the  remaining  data  being  as  before,  compute  (a)  the 
acceleration  of  each  particle  and  (b)  the  acceleration  of  the  mass- 
center.  Ans.  (b)  1 . 5  ft. -per-sec. -per-sec. 

3.  Two  bodies  of  masses  12  lbs.  and  8  lbs.  respectively,  con- 
nected by  an  elastic  string,  are  projected  in  any  manner  and  left  to 


MOTION   OF    ANY   SYSTEM    OF    PARTICLES.  325 

the  action  of  gravity.  At  a  certain  instant  the  bodies  are  moving 
horizontally  in  opposite  directions,  each  at  the  rate  of  20  ft.-per-sec. 
Determine  the  subsequent  motion  of  the  mass-center. 

Ans.  Its  motion  will  be  that  of  a  projectile  having  initially  a  hori- 
zontal velocity  of  4  ft.  -per-sec. 

4.  Take  data  as  in  Ex.  3,  except  that  the  velocity  of  the  mass  of 
12  lbs.  is  20  ft.-per-sec.  and  that  of  the  mass  of  8  lbs.  is  30  ft.-per- 
sec.  in  the  opposite  direction,  (a)  Determine  the  motion  of  the 
mass-center,  (b)  If  the  velocities  are  vertical  instead  of  horizontal, 
how  will  the  solution  be  changed  ? 

Ans.  (a)  Its  motion  will  be  that  of  a  body  falling  from  rest. 

5.  Show  that  the  center  of  mass  of  any  system  of  particles  will 
move  uniformly  in  a  straight  line  if  no  external  force  acts  upon  any 
particle,  or  if  the  vector  sum  of  all  the  external  forces  is  zero. 


§  2.  Angular  Motion  of  a  Material  System. 

381.  Angular  Motions  of  Individual  Particles. — From  the  first 
pair  of  equations  (2)  (Art.  378)  may  be  obtained  the  following : 

Yxxx  —  Xxyx  =  mx(xxyx—yxxx).         .         .     (1) 

The  second  member  of  this  equation  may  be  written  in  the  equiv- 
alent form 

~\mx(xxyx  —  yxxx)\ 
at 

as  may  be  verified  by  differentiation.  This  expression  has  a  simple 
meaning.  Since  mxxx  and  Mxyl  are  the  axial  components  of  the 
momentum  of  the  particle,  the  moment  of  this  momentum  with  re- 
spect to  the  origin  of  coordinates  is  mx(xxyx  —  y^x)-  Representing 
this  by  HXi  the  second  member  of  equation  (1)  is  equal  to  dHJdt. 
Since  XA  and  Yx  are  the  axial  components  of  the  resultant  of  all 
forces  acting  upon  the  particle,  the  first  member  of  equation  (1)  is 
equal  to  the  sum  of  the  moments  of  these  forces  with  respect  to  the 
origin  of  coordinates.  Representing  this  by  Lx,  the  equation  may 
be  written  in  the  form 

Lx  =  dHJdt. 

Let  a  similar  equation  be  written  for  every  particle ;  then  by 
addition, 

£1  +  Z»:+    :    .    .     =  dHJdt  +  dHJdt  +     ....     (2) 


326  THEORETICAL    MECHANICS. 

382.  Elimination  of  Internal  Forces. —  The  first  member  of 
equation  (2)  is  the  sum  of  the  moments  of  all  forces,  external  and 
internal,  acting  upon  the  particles  of  the  system.  But  the  sum  of 
the  moments  of  the  internal  forces  is  zero,  because  the  two  forces 
acting  between  two  particles  are  not  only  equal  and  opposite  but  are 
collinear,  so  that  their  moments  are  equal  in  magnitude  and  opposite 
in  sign.  If,  therefore,  the  sum  of  the  moments  of  the  external  forces 
is  denoted  by  L, 

L  =  LX  +  L2+     .     .     .     . 

383.  Equation  of  Angular  Motion  for  the  System. — Let  the 

total  moment  of  momentum  of  the  system  (i.  e.,  the  sum  of  the  mo- 
ments of  the  momenta  of  the  individual  particles)  be  represented  by 

H.    Then 

////,-.//  .     .     .     , 

and  equation  (2)  reduces  to  the  form 

L  =  dHldt        ....     (3) 

In  deducing  this  equation,  the  origin  of  moments  was  taken  at 
the  origin  of  coordinates ;  but  this  may  be  any  point  in  the  plane  of 
the  motion. 

384.  Principle  of  Angular  Momentum. —  The  sum  of  the  mo- 
ments of  the  momenta  of  all  the  particles  is  called  the  angular 
momentum  of  the  system  (Art.  327).  Equation  (3)  expresses  the 
proposition  that 

The  rate  of  change  of  the  angular  momentum  of  a  system  about 
any  point  is  equal  to  the  sum  of  the  moments  of  the  external  forces 
about  that  point. 

The  following  are  important  special  cases  of  this  general  prin- 
ciple : 

If  no  external  force  acts  upon  any  particle,  the  angular  momen- 
tum of  the  system  about  every  point  remains  'constant. 

If  the  resultant  of  the  external  forces  acts  always  through  a 
fixed  point,  the  angular  momentum  with  respect  to  that  point  remains 
constant. 

The  above  equation  of  angular  motion,  together  with  the  equa- 
tions of  motion  of  the  center  of  mass,  express  all  that  can  be  deter- 
mined regarding  the  motion  of  a  material  system  from  the  external 
forces  alone. 


MOTION    OF    ANY   SYSTEM    OF    PARTICLES.  327 

§  3.  Effective  Forces ;  D'  Alemberf  s  Principle. 

385.  Effective  Forces  Defined. —  The  resultant  of  all  forces 
(both  external  and  internal)  acting  upon  any  particle  is  called  the 
effective  force  for  that  particle.  Such  resultants  for  all  the  particles 
of  a  system  make  up  the  effective  forces  for  the  system. 

For  a  particle  of  mass  tn  and  acceleration  pt  the  effective  force 
has  the  magnitude  mp,  its  direction  being  that  of  the  acceleration  p. 
Therefore,  if  the  position,  mass  and  acceleration  of  every  particle  of 
the  system  are  known,  the  effective'  forces  are  completely  known. 
They  are,  in  fact,  forces 

^,A,  f*%p%\  **tAi    •    •    v    1 

having  directions  identical  with  those  of  pA ,  p2 ,  p3 ,  .  .  .  ,  and 
points  of  application  coincident  with  the  positions  of  the  particles. 
The  magnitude,  direction  and  line  of  action  of  the  effective  force 
acting  upon  any  moving  particle  are  in  general  continuously  changing. 

386.  Resultant  Effective  Force. —  If  all  the  effective  forces  for 
a  system  be  combined  by  the  methods  used  in  combining  forces  in 
the  statics  of  a  rigid  body,  the  resultant  may  be  called  the  resultant 
effective  force  for  the  system. 

For  coplanar  forces  this  resultant  is  a  single  force  or  a  couple. 

387.  Relation  Between  Effective  Forces  and  External  Forces. — 

Since  the  effective  force  for  any  particle  is  the  resultant  of  the  exter- 
nal and  internal  forces  acting  upon  that  particle,  the  resultant  effective 
force  for  the  system  may  be  found  by  combining  all  external  and 
internal  forces  acting  upon  all  particles  of  the  system.  But  the  re- 
sultant of  all  the  internal  forces,  combined  in  the  prescribed  manner, 
is  zero,  since  the  entire  system  of  internal  forces  is  made  up  of 
stresses.     It  follows  that 

The  resultant  of  the  effective  forces  is  equal  in  all  respects  to  the 
resultant  of  the  external  forces. 

It  should  be  observed  that  the  word  resultant  is  here  used  in  an 
arbitrary  sense.  To  say  that  a  force  is  the  resultant  of  other  forces 
means  strictly  that  it  is  equivalent  to  them  in  effect.  Two  or  more 
forces  applied  to  different  free  particles  are  not  equivalent  to  any 
single  force.  It  is,  however,  convenient  to  use  the  term  resultant 
whenever  it  is  desired  to  express  the  fact  that  several  forces  are 


328  THEORETICAL    MECHANICS. 

combined  according  to  the  same  rules  as  in  the  statics  of  a  rigid 
body. 

The  meaning  of  the  proposition  stated  above  is  that  the  system 
of  effective  forces  and  the  system  of  external  forces  satisfy  exactly 
the  same  conditions  which  are  satisfied  by  two  eqtiivalent  systems  of 
forces  applied  to  the  same  rigid  body. 

This  principle  leads  immediately  to  algebraic  equations  of  two 
forms:  resolution  equations  and  moment  equations.     Thus, 

(a)  The  sum  of  the  resolved  parts  of  the  external  forces  in  any 
direction  is  equal  to  the  sum  £>f  the  resolved  parts  of  the  effective 
forces  in  the  same  direction. 

{b)  The  sum  of  the  moments  of  the  external  forces  about  any 
axis  is  equal  to  the  sum  of  the  moments  of  the  effective  forces  about 
that  axis. 

If  the  motion  is  restricted  to  a  plane,  three  independent  equa- 
tions may  be  written,  of  which  at  least  one  must  be  a  moment  equa- 
tion. For  motion  in  three  dimensions  six  independent  equations  may 
be  written.* 

388.  Equations  of  Linear  and  Angular  Motion.  —  The  prin- 
ciple that  the  external  forces  and  the  effective  forces  form  equivalent 
systems  leads  immediately  to  the  equations  of  motion  of  the  mass- 
center  given  in  Art.  380,  and  to  the  equation  of  angular  motion  given 
in  Art.  383.  The  former  are  obtained  by  resolving  along  the  coordi- 
nate axes,  the  latter  by  taking  moments  about  the  origin  of  coor- 
dinates. 

The  axial  components  of  the  effective  force  for  the  particle  mx 
are  obviously  mix1  and  mxyXi  and  the  sums  of  the  axial  components 
for  all  the  particles  are 

mlxl  -j-  tn%xt  -}-...=  (mx  +  m2  -\-    .     .    .   )(^2^/<^/2), 

m\y\  ~\~  m%y>  +'...=  (*#!  -f-  m.,  -f-     .     .     .   ){d2yldt2). 

Equating  these  respectively  to  X  and  Y,  the  axial  components  of 
the  resultant  external  force,  the  resulting  equations  are  those  before 
found  for  the  motion  of  the  mass- center. 

Again,  the  moment  of  the  effective  force  for  the  particle  mx  is 
mx{xxyx  —  yxx^t 

*  These  statements  may  be  proved  by  reasoning  similar  to  that  employed 
in  Arts.  104  and  173. 


MOTION    OF    ANY    SYSTEM    OF    PARTICLES.  329 

which,  as  shown  in  Art.  381,  is  equal  to  dHJdt,  if  Hx  is  the  angular 
momentum  of  vix  about  the  origin ;  and  the  sum  of  the  moments  of 
all  the  effective  forces  is  therefore  dHjdt,  if  H  is  the  angular  mo- 
mentum of  the  system.  Equating  this  to  Z,  the  moment  of  the 
resultant  external  force,  the  result  is  the  equation  of  angular  motion 
given  in  Art.  383. 

389.  System  of  Particles  Rigidly  Connected. — In  general  the 
three  independent  equations,  free  from  the  internal  forces,  which 
may  be  written  in  accordance  with  the  general  principle  of  Art.  387, 
are  insufficient  to  determine  the  motion  completely.  But  if  the  par- 
ticles are  so  connected  that  their  motions  are  compelled  to  satisfy 
certain  geometrical  conditions,  these  conditions,  together  with  the 
three  dynamical  equations,  may  suffice  to  determine  the  motion  of 
every  particle. 

An  important  case  is  that  in  which  the  particles  are  rigidly  con- 
nected, so  that  the  distance  between  any  two  particles  remains  in- 
variable. It  will  be  seen  in  the  following  chapters  that  the  motion  of 
such  a  system  is  completely  determined  by  the  three  dynamical  equa- 
tions, so  that  the  motion  of  a  rigid  system  may  be  determined  from 
the  external  forces  alone. 

390.  D'Alembert's  Principle. —  The  principle  of  Art.  387,  when 
applied  to  a  rigid  body,  is  known  as  D'Alembert's  principle.  It  is 
the  basis  of  all  rules  for  the  solution  of  problems  relating  to  the  mo- 
tions of  such  bodies.  The  foregoing  discussion  makes  it  clear  that 
the  principle  is  equally  true  for  non-rigid  bodies  or  systems.  But  it 
is  only  in  the  case  of  a  rigid  system  that  the  principle  suffices  for  the 
complete  determination  of  the  motion. 

391.  Motion  in  Three  Dimensions. —  Although  much  of  the 
foregoing  discussion  has  referred  to  two-dimensional  motion,  most 
of  the  results  may  easily  be  extended  to  the  case  of  motion  in  three 
dimensions. 

The  general  principle  of  the  equivalence  of  the  external  forces 
and  the  effective  forces  (Art.  387)  is  obviously  unrestricted.  From 
this  follow  at  once  the  equations  of  motion  for  the  mass-center,  which 
will  be  three  in  number  in  the  general  case,  and  which  are  equiva- 
lent to  the  general  proposition  stated  at  the  end  of  Art.  380.  The 
equation  of  angular  motion  is  also  true  for  three-dimensional  motion, 
and  may  be  written  in  the  same  form  as  in  Art.  383  ;  but  moments 


330  THEORETICAL    MECHANICS. 

must  be  taken  about  an  axis  instead  of  a  point,  and  by  taking  mo- 
ments about  three  axes  three  independent  equations  may  be  obtained. 
Thus,  for  three-dimensional  motion  three  independent  equations  of 
resolution  may  be  written,  and  also  three  independent  moment- 
equations.  In  the  ease  of  a  rigid  body  these  six  independent  equa- 
tions suffice  to  completely  determine  the  motion. 

392.  Moment  of  Inertia. —  In  the  theory  of  the  motion  of  a 
rigid  body  a  certain  quantity  called  the  moment  of  inertia  plays  an 
important  part.  The  following  chapter  will  therefore  include  a  dis- 
cussion of  this  quantity  as  a  necessary  preliminary  to  the  study  of 
the  motion  of  a  rigid  body. 


CHAPTER  XIX. 

MOMENT    OF    INERTIA. 

§  i.  Moment  of  Inertia  of  a  Rigid  Body. 

393.  Definition. —  The  moment  of  inertia  of  a  body  with  respect 
to  any  axis  is  the  sum  of  the  products  obtained  by  multiplying  every 
elementary  mass  by  the  square  of  its  distance  from  the  axis. 

The  value  of  the  quantity  thus  denned  is  seen  to  depend  upon 
the  size  and  shape  of  the  body,  the  distribution  of  its  mass,  and  the 
position  of  the  axis.  For  a  given  body  the  moment  of  inertia  with 
respect  to  a  line  fixed  in  the  body  is  a  definite  constant ;  for  different 
axes,  the  moment  of  inertia  will  have  different  values. 

394.  Moment  of  Inertia  of  a  System  of  Discrete  Particles. — 

For  a  system  of  discrete  particles  of  finite  mass,  the  value  of  the 
moment  of  inertia  may  be  computed  by  the  formula 

/=  mxrf  +  m2r*  -(-...=  2/^r2,  .     (1) 

in  which  mx ,  mt ,  .  .  .  denote  the  masses  of  the  particles,  rx , 
r,i  .  .  .  their  respective  distances  from  the  assumed  axis,  and 
/  the  moment  of  inertia  with  respect  to  that  axis. 

395.  Moment  of  Inertia  of  Continuous  Mass. —  In  computing 
moments  of  inertia,  bodies  are  assumed  to  consist  of  matter  distrib- 
uted continuously  throughout  space,  so  that  any  finite  mass  occupies 
a  finite  volume,  and  the  moment  of  inertia  is  in  general  found  by 
integration. 

An  approximate  value  of  the  moment  of  inertia  of  a  continuous 
mass  may  be  obtained  as  follows : 

Let  M  denote  the  whole  mass  of  the  body,  and  let  it  be  divided 
into    small    portions    whose    masses    are    AM~l,   A^f2,      . 
Let  rlt  rti     .     .     .     denote  the  distances  from  the  assumed  axis  to 
any  definite  points  of  the  elementary  masses.      The  value  of  the  mo- 
ment of  inertia  is  approximately 

f=Tr*AMl  +  r*AMt+    .    .    .    , 

the  approximation  being  closer  the  smaller  the  elements  of  mass  are 
taken. 


332  THEORETICAL    MECHANICS. 

An  exact  value  of  the  moment  of  inertia  is  found  by  determining 
the  limit  of  the  approximate  value  as  all  the  elementary  masses  are 
made  to  approach  zero,  their  number  being  increased  without  limit. 
If  the  variable  r  denotes  the  distance  of  the  differential  element  of 
mass  dM  from  the  assumed  axis,  the  exact  value  of  the  moment  of 
inertia  is 

I=fr2dM,  .         .  .  .     (2) 

the  limits  of  the  integration  being  so  assigned  as  to  include  the  entire 
body. 

Choice  of  element  of  mass. — The  differential  element  of  mass  may 
be  of  the  first,  second  or  third  order ;  and  the  integration  indicated 
in  equation  (2)  may  be  either  single,  double  or  triple,  depending 
upon  the  order  of  the  differential  element.  The  element  must  be 
so  taken  that  all  points  in  it  have  a  common  value  of  r. 

396.  Radius  of  Gyration. — The  radius  of  gyration  of  a  body 
with  respect  to  any  axis  is  the  distance  from  the  axis  at  which  a 
single  particle,  of  mass  equal  to  that  of  the  body,  must  be  located  in 
order  that  its  moment  of  inertia  may  be  equal  to  that  of  the  body. 

If  M  denotes  the  total  mass  of  the  body,  /  its  moment  of  inertia 
with  respect  to  any  axis,  and  k  its  radius  of  gyration  with  respect  to 
the  same  axis,  we  have 

I=Mk2)     kl  =  I\M. 

from  which  k  can  be  computed  when  /  is  known. 

If  the  whole  mass  M  be  subdivided  into  parts  Mx,  M2, 
.  .  .  ,  whose  radii  of  gyration  are  kx ,  k., ,  .  .  .  ,  the  mo- 
ment of  inertia  of  M  is 

I=zMxk?  +  M&  4-    .... 

It  may  be  possible  to  choose  a  differential  element  of  M  which  has 
one  or  two  finite  dimensions,  but  whose  radius  of  gyration  about  the 
given  axis  is  known.  If  dM  is  such  an  element  and  r'  its  radius  of 
gyration, 

I=fr'*dM.         ....     (3) 

This  agrees  in  form  with  equation  (2),  and  in  fact  reduces  to  (2) 
when  all  points  of  dM  are  equally  distant  from  the  axis,  since  in  that 
case  r'  =  r. 

397.  Units  Involved  in  Moment  of  Inertia. —  Each  term  in 
the  series  whose  sum  is  equal  to  the  moment  of  inertia  is  the  product 


MOMENT    OF    INERTIA.  333 

of  a  mass  into  the  square  of  a  length.  The  unit  in  which  moment  of 
inertia  is  expressed  is  therefore  of  dimensions  ML*.  In  using  the 
value  of  the  moment  of  inertia  in  the  formulas  of  dynamics,  care  must 
be  observed  that  proper  units  of  mass  and  length  are  used  in  com- 
puting it.  The  proper  units  to  be  used  in  any  case  may  be  seen 
from  the  nature  of  the  particular  application.  In  the  present  dis- 
cussion of  general  principles  and  of  methods  of  computing  the  moment 
of  inertia,  any  units  of  mass  and  length  may  be  understood. 

It  is  usually  an  aid  in  keeping  the  units  clearly  in  mind  to  express 
the  moment  of  inertia  as  the  product  of  the  mass  into  the  square  of 
the  radius  of  gyration.  The  latter  quantity  is  obviously  a  simple 
linear  magnitude. 

398.  Body  of  Uniform  Density. —  Let  V  denote  the  whole  vol- 
ume and  M  the  whole  mass  of  a  body,  and  let  p  be  the  density  at 
any  point.     Then 

dM=  pdV. 

If  the  density  is  uniform  throughout  the  body,  equation  (2)  becomes 

I^pffdV.       .        .        .        .    (4) 

The  method  of  applying  this  formula  is  shown  in  the  solution  of  some 
of  the  following  examples. 

Examples. 

1 .  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cylinder  with  respect  to  its  axis  of  figure. 

Let  /  denote  the  length  of  the  cylinder,  a  the  radius  of  the  base, 
p  the  density.  Taking  any  right  section,  let  r,  0  be  the  polar  coor- 
dinates of  any  point  in  the  cross-section,  the  pole  of  the  system  of 
coordinates  being  at  the  center.  If  dA  is  a  differential  element  of 
the  area  of  the  cross-section,  we  may  take  as  the  element  of  mass  a 
differential  prism  of  cross-section  dA  and  length  /.  The  general 
formula  for  the  moment  of  inertia  becomes 

I=fr*dM=fr*(pldA)  =  ptfr'dA. 
Expressing  dA  in  terms  of  the  coordinates  r,  6y  its  value  is  rdOdr\ 
whence  ^2tz  r»a 

l=pl\       I    r*d6dr. 

•/ o      Jo 

Either  integration  may  be  performed  first,  since  the  limits  for  each 
are  constants.     The  result  of  the  double  integration  between  the 
designated  limits  is       /=  ^dlpfe  =  Ma2/2, 
since  the  total  mass  M  is  equal  to  irdllp. 


334  THEORETICAL    MECHANICS. 

If  k  is  the  radius  of  gyration, 

1=  Mk2  =  Ma2/ 2  ;     k2  =  a2/2. 

The  value  of  the  radius   of  gyration  is  thus  independent  of  the 
density. 

The  differential  element  may  be  so  chosen  that  only  a  single  in- 
tegration is  needed.  Thus,  if  the  element  is  taken  as  the  mass 
included  between  two  cylindrical  surfaces  of  radii  r  and  r  -\-  dr, 
dM  =  2/pir  r  dr,  and 

I=Cr2dM=  27r/pf  rAdr  =  iraHp^, 
as  before.  *  ° 

2.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  parallelopiped  with  respect  to  an  axis  of  symmetry. 

Ans.  If  a,  b  are  the  lengths  of  the  edges  perpendicular  to  the 
axis,  k2=  O2  +  0*)/l2. 

3.  What  is  the  radius  of  gyration  of  a  homogeneous  square  prism 
whose  dimensions  in  inches  are  4X4X8,  with  respect  to  an  axis 
through  the  centroids  of  the  square  sections  ?  How  does  the  value 
of  the  radius  of  gyration  depend  upon  the  density  ?  How  upon  the 
length?  Ans.  k=  1.6331ns. 

4.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
body  composed  of  two  coaxial  homogeneous  cylinders,  the  diameters 
of  the  circular  sections  being  4  ins.  and  8  ins.  respectively,  and  the 
masses  8  lbs.  and  6  lbs. 

Ans.  k2  =  32/7,  /=  64,  the  inch  and  pound  being  the  units  of 
length  and  mass. 

5.  A  body  is  made  up  of  two  portions  of  masses  J^  and  M2, 
their  radii  of  gyration  with  respect  to  a  certain  axis  being  kx  and  k2 . 
Determine  the  moment  of  inertia  and  the  radius  of  gyration  of  the 
whole  body  with  respect  to  the  same  axis. 

Ans.  I  =  Mxk2  +  M,k2.     kl  =  (Mxk2  +  M,k2)j{,Mx  -f  M2). 

6.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  sphere  with  respect  to  an  axis  through  the  center. 

Let  a  be  the  radius  of  the  sphere  and  p  its  density,  and  let  v 
denote  the  radius  of  a  circular  section  perpendicular  to  the  axis  and 
distant  z  from  the  center.  Let  dM  be  the  mass  of  an  element  in- 
cluded between  this  section  and  one  parallel  to  it  at  distance  dz. 
The  moment  of  inertia  of  dM  is  (by  Ex.  1) 

\v2dM=  \v\irv2pdz)  =  \-rrpv'dz, 

and  the  moment  of  inertia  of  the  sphere  is  found  by  integrating  this 
differential  expression  between  limits  z  =  —  a  and  z  =  -\-  a.     That  is, 

7  =  f    iirpv*ds  =  \irp  f    (a2  —  z2)2dz  =  87^715. 

J  —  a  J —a 


MOMENT   OF    INERTIA.  335 

Since  M,  the  mass  of  the  sphere,  is  ^Trpa3/^,  the  value  of  the 
moment  of  inertia  is  2Ma2/s>  and  therefore 

k2  =  2^/5. 

7.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cone  with  respect  to  its  geometrical  axis. 

Let  a  be  the  radius  of  the  base,  Ji  the  altitude,  M  the  mass  and 
p  the  density.  Let  v  be  the  radius  of  the  circular  section  distant  z 
from  the  vertex,  and  let  dM  be  the  elementary  mass  included  be- 
tween this  section  and  another  distant  dz  from  it.  The  moment  of 
inertia  of  this  element  is  (by  Ex.  i) 

dl=  iv2dM=  \v\irv2dz)p  ===  \irprfdz. 
Integrating, 

v*dz  =53  \irp  I    (a^/k^dz  =  ira'/ip/ 10. 

O  •'o 

Since  M  =  ira^hpl^,  we  may  write 

/  =  (y?/io)M;     .  • .     k2  =  3^/10. 

8.  Determine  approximately  the  moment  of  inertia  and  radius  of 
gyration  of  a  homogeneous  circular  lamina  of  small  uniform  thickness 
with  respect  to  an  axis  through  the  centroid  parallel  to  the  circular 
faces. 

Taking  a  circular  section  of  the  lamina  through  the  inertia-axis, 
let  r,  6  be  the  polar  coordinates  of  any  point  in  this  section,  the  pole 
being  at  the  center  and  the  initial  line  coinciding  with  the  inertia- 
axis.  Let  a  be  the  radius  of  the  circle,  h  the  thickness  of  the  lamina 
and  p  its  density.  Take  as  element  of  mass  a  differential  prism  of 
altitude  h  and  cross-section  rdQdr\  the  moment  of  inertia  of  this 
element  is  approximately  (r  sin  Q)\phrdQ  dr).  The  required  mo- 
ment of  inertia  is  therefore,  approximately, 

/  ==  f27r  Cphr*  sin2  0  dOdr=  irrfAp/4. 

V  o      J  o 

If  k  is  the  radius  of  gyration, 

k1  =  Hirdlhp  =  a2 1 4  ;     .  • .     k  =  a/ 2. 

9.  Determine  the  radius  of  gyration  of  a  spherical  shell  of  uni- 
form small  thickness  and  uniform  density  with  respect  to  a  diameter. 

Ans.  If  a  is  the  radius,  k2  =  \d2. 

10.  Determine  the  radius  of  gyration  of  a  spherical  shell  of  uni- 
form density  and  of  any  thickness  with  respect  to  a  diameter. 

Ans.  If  the  outer  and  inner  radii  are  a  and  b,  k2  — 
2O5  —  ^)/5(«3—  P). 

399.  Body  of  Variable  Density. — If  the  density  is  not  uniform 
throughout  the  body,  the  factor  p  cannot  be  placed  before  the  sign 
of  integration.     The  formula  for  /  takes  the  form 


336  THEORETICAL    MECHANICS. 

I  =  fr'idM  =  fr2pdV.  .  .  .  (5) 
The  integration  cannot  be  effected  unless  the  law  of  variation  of  p 
throughout  the  body  is  known. 

Examples. 

1.  Compute  the  radius  of  gyration  of  a  right  circular  cylinder  with 
respect  to  its  axis  of  figure,  assuming  the  density  to  be  given  by  the 
equation  p  =  p0  -\-  cr,  r  being  the  distance  from  the  axis  and  c  a 
constant. 

Proceeding  as  in  Ex.  1,  Art.  398,  we  may  write  at  once 

1=1)       (    pr'  dOdr  =  I J       )     (p0  +  cr)rzdddr. 

Integrating  and  reducing, 

/=  7r/(5/v*4  +  4^5)/io  =  iraH(y\px  -f  f?0)/io, 

px  being  the  density  at  the  outer  surface. 

The  value  of  M  may  be  found  by  integration  : 

M=7rdz/(2Pl  +  />0)/3. 

Hence  #  =  I/M  =  [(4^  +  Po)/(2Pl  +  pjfop/io). 

If  p0  =  p{  =  p,  these  values  reduce  to  those  found  for  the  case 
of  uniform  density. 

2.  Determine  the  moment  of  inertia  of  a  sphere  with  respect  to  a 
diameter,  assuming  the  density  to  vary  directly  as  the  distance  from 
the  center,  being  zero  at  the  center  and  px  at  the  surface. 

Ans.   M  =  vc^px\     k2  =  4a2 lg. 

400.  Relation  Between  Moments  of  Inertia  With  Respect  to 
Parallel  Axes. —  The  moment  of  inertia  of  a  body  with  respect  to 
any  axis  is  equal  to  its  moment  of  inertia  with  respect  to  a  parallel 
axis  through  the  center  of  mass  plus  the  product  of  the  whole  mass 

into  the  square  of  the  distance  between 
-.^^  the  two  axes. 

c    n     ;>  Let  Fig.  162  represent  a  section  of 

y  ■     \  the  body  by  a  plane  perpendicular  to 

j      )         the  assumed   axes,    the  point   0  rep- 
0'     i  O  X      resenting   the  central  axis  and   O'  an 

\^» — . J  axis  distant  a  from   the  central  axis. 

Fig.  162.  Let  0'  OX  be  the  trace  of  a  plane  con- 

taining both  axes  ;  OY  and  O'  Y'  the 
traces  of  planes  perpendicular  to  O' OX.  Let  x,  y  be  the  perpen- 
dicular distances  of  any  point  of  the  body  from  the   planes  OY 


MOMENT   OF    INERTIA.  337 

and  OX  respectively,  the  distances  of  the  same  point  from  O'Y' 
and  O ' X  being  therefore  x  +  a  and  y. 

Let  /  denote  the  moment  of  inertia  of  the  body  with  respect  to 
the  axis  0,  and  /'  the  moment  of  inertia  with  respect  to  the  axis  O'. 
Then 

/  =  /(*■  +  y2)dM; 

/'  =f[(x  +  a)2  +  f]dMr=f{x2  +y>  +  a2  +  2**^CAf 
=  /(*2  +  >*M#f  +  tffdM  +  2afxdM, 

the  limits  of  integration  in  every  case  being  so  taken  as  to  include 
the  whole  body. 

In  the  final  expression  for  T\  the  first  term  is  equal  to  /  and  the 
second  to  a2M.  The  third  term  is  equal  to  zero,  since  the  plane 
OY  (from  which  x  is  measured)  contains  the  center  of  mass.  (See 
Art.  159.)     Hence 

I'  =  I+Ma2y       .         .         .         .     (1) 

and  the  proposition  is  proved. 

Relation  between  radii  of  gyration. —  If  k  and  k'  are  the  radii 
of  gyration  of  the  body  with  respect  to  the  axes  O  and  0 '  respect- 
ively, /  =  Mk\  F  =  Mk"\  and  therefore 

k'2  =  k2  +  a\        .  .         .         .     (2) 

Examples. 

1.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cylinder  with  respect  to  an  axis  coin- 
ciding with  one  of  the  straight  elements  of  the  surface. 

Ans.  If  a  =  radius  of  circular  section,  k2  =  3a2/ 2. 

2.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  square  prism  with  respect  to  an  axis  coinciding  with 
one  of  the  edges.  Apply  the  result  to  a  prism  whose  mass  is  1 2  lbs. 
and  whose  linear  dimensions  in  inches  are  4X4X8. 

Ans.  If  the  axis  coincides  with  one  of '  the  longer  edges,  k  = 
(8v/6)/3. 

3.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cone  whose  altitude  is  6  ft.  and  the  radius 
of  whose  base  is  2  ft. ,  with  respect  to  an  axis  parallel  to  the  geo- 
metrical axis  and  2  ft.  from  it ;  the  whole  mass  being  M  lbs 

Ans.  k2=  5.2  ft.2 

4.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cylinder  with  respect  to  an  axis  through 
the  center  of  mass  coinciding  with  the  diameter  of  a  right  section. 


338  THEORETICAL    MECHANICS. 

Let  /  be  the  length  of  the  cylinder,  a  the  radius  of  the  base,  and 
p  the  density.  Let  dM  be  the  mass  included  between  two  circular 
sections  distant  x  and  x  -\-  dx  from  the  center  of  mass.  The  radius 
of  gyration  of  dM  with  respect  to  a  diameter  is  aJ2  ;  with  respect  to 
a  parallel  axis  through  the  centroid  of  the  cylinder  the  square  of  its 
radius  of  gyration  is  therefore  x2  -\-  a2/ '4,  and  its  moment  of  inertia  is 

dl=  (x2  +  a2/i)dM=  (x2  +  a2/4.)(7ra2pdx). 
Integrating  between  limits  x  =  —  I/2  and  x  =  //2, 
I=ira2lp{2>a2  +  /2)/i2; 
k2=^a2  +  /2)/i2. 

5.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cylinder  with  respect  to  an  axis  coin- 
ciding with  a  diameter  of  the  base.     Ans.  k2  =  (3a2  -+-  4/2)/i2. 

6.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
homogeneous  right  circular  cone  with  respect  to  an  axis  through  the 
vertex  perpendicular  to  the  geometrical  axis. 

Ans.  If  h  is  the  altitude  and  a  the  radius  of  the  base,  k2  = 
3(/z2  +  «74)/5. 

401.  Product  of  Inertia.  —  The  sum  of  the  products  obtained 
by  multiplying  every  elementary  mass  of  a  body  by  the  product  of 
its  distances  from  two  planes  perpendicular  to  each  other  is  called 
the  product  of  inertia  of  the  body  with  respect  to  those  planes. 

For  a  system  of  discrete  particles,  let  mlt  m.2f  .  .  .  be  the 
masses  of  the  particles,  xiy  x2  .  .  .  their  distances  from  one 
plane  and  yx ,  y2 ,     .  their  distances  from  the  other  ;  and  let 

H  denote  the  product  of  inertia.     Then 

N==  mxxxyx  +  m2x2y2  +     .     .  '  .     =  Imxy.         .     (1) 

For  a  continuous  mass,  let  x  and  y  denote  the  distances  of  an 
element  of  mass  dM  from  the  two  planes  ;  then 

H^fxydM.         ....     (2) 

Examples. 

1.  Compute  the  product  of  inertia  of  a  homogeneous  rectangular 
parallelopiped  with  respect  to  planes  containing  two  intersecting 
faces. 

The  three  linear  dimensions  of  the  body  being  a,  b  and  c,  let  the 
product  of  inertia  be  found  with  respect  to  two  intersecting  faces 
parallel  to  the  dimension  c.  The  traces  of  these  planes  are  repre- 
sented by  OX  and  (9F(Fig.  163).  The  density  being  p,  the  ele- 
ment of  mass  dM  may  be  taken  equal  to  pcdxdy,  and 


MOMENT    OF    INERTIA.  339 


j     pcxydxdy  ==  pcftfq^  =  Mabj 4. 

o     «^0 


dM 

X 


H  = 

2.  Compute  the  product  of  inertia  of  a  homogeneous  right  cir- 
cular cylinder  with  respect  to  a  plane  containing  one  of  the  bases  and 
a  plane  tangent  to  the  cylindrical  surface. 

An s.   If  a  =  radius  of  base  and  /  = 
altitude,  H  =  Malh. 

3.  Compute  the  product  of  inertia  of 
a  homogeneous  right  circular  cylinder 
with  respect  to  a  plane  containing  the 
geometrical  axis  and  any  plane  perpen-       q~ 
dicular  to  that  axis.      Ans.  H  =  o.  Fig.  163. 

402.  Product  of  Inertia  of  a  Body  Having  a  Plane  of  Sym- 
metry.—  The  product  of  inertia  of  a  body  of  uniform  density  with 
respect  to  two  planes,  one  of  which  is  a  plane  of  symmetry,  is  zero. 
For,  let  x  be  the  distance  of  any  point  from  the  plane  of  symmetry 
and  y  its  distance  from  the  other  plane.  Corresponding  to  any  ele- 
ment of  mass  for  which  x  =  a,  y  =  b,  there  is  an  equal  element  for 
which  x  =  —  a,  y  =  b.  The  sum  of  the  products  of  inertia  of  such 
a  pair  of  elements  is  zero,  hence  the  product  of  inertia  of  the  whole 
mass  is  zero. 

§  2.  Moment  of  Inertia  of  a  Plane  Area. 

403.  Definition. —  In  the  theory  of  the  strength  and  elasticity  of 
beams,  columns  and  shafts,  there  is  involved  a  quantity  whose  value 
is  given  by  an  expression  analogous  in  form  to  the  expression  for  the 
moment  of  inertia  of  a  solid  body.  The  importance  of  this  quantity, 
and  its  analogy  to  the  moment  of  inertia  as  above  denned,  make  it 
desirable  to  give  it  attention  here.     It  may  be  defined  as  follows : 

The  moment  of  inertia  of  a  plane  area  with  respect  to  any  axis  is 
the  sum  of  the  products  obtained  by  multiplying  every  elementary 
area  by  the  square  of  its  distance  from  the  axis. 

In  the  most  important  applications  the  axis  is  taken  either  in  the 
plane  of  the  given  area  or  perpendicular  to  it.  The  moment  of  in- 
ertia of  a  plane  area  with  respect  to  an  axis  perpendicular  to  its  plane 
is  called  a  polar  moment  of  inertia. 

The  term  radius  of  gyration  is  also  used  in  connection  with  areas, 
being  defined  by  the  equation 

MP  =  L 


34°  THEORETICAL    MECHANICS. 

in  which  /  is  the  moment  of  inertia  of  the  area,  k  its  radius  of  gyra- 
tion and  M  the  total  area. 

It  is  obvious  from  the  definition  that  the  moment  of  inertia  of  a 
plane  area  may  be  computed  by  a  method  similar  in  every  respect  to 
that  used  in  case  of  a  mass,  except  that  elements  of  area  must  be  used 
instead  of  elements  of  mass. 

It  may  be  noticed  also  that  the  proposition  of  Art.  400  regarding 
the  relation  between  moments  of  inertia  with  respect  to  parallel  axes 
holds,  using  total  area  instead  of  total  mass. 

Examples. 

1 .  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
circular  area  with  respect  to  an  axis  through  the  center  perpendicular 
to  the  plane. 

[Follow  the  method  employed  in  solving  Ex.  1,  Art.  398,  but  use 
the  element  of  area  rd6  dr  instead  of  the  element  of  mass  plrdd  dr. 
The  radius  of  gyration  of  a  circular  section  of  the  cylinder  has  the 
same  value  as  that  of  the  cylinder.  ] 

2.  Show  that  the  radius  of  gyration  of  a  circular  area  with  respect 
to  a  diameter  is  half  the  radius  of  the  circle. 

3.  If  b  and  h  are  the  sides  of  a  rectangle,  prove  that  its  moment 
of  inertia  with  respect  to  an  axis  containing  the  centroid  and  parallel 
to  the  side  b  is  b/13/ 12.  From  this  result  and  the  proposition  of  Art. 
400  deduce  the  value  of  the  moment  of  inertia  with  respect  to  an 
axis  coinciding  with  a  side  of  the  rectangle.  Ans.  k2  =  //2/3- 

4.  Compute  the  moment  of  inertia  of  a  triangular  area  of  base  b 
and  altitude  h  with  respect  to  an  axis  containing  the  vertex  and  par- 
allel to  the  base.  Ans.  I  =  bfffa  ;     k1  =  k2/2. 

5.  From  the  result  of  Ex.  4  and  the  proposition  of  Art.  400,  de- 
termine the  moment  of  inertia  and  radius  of  gyration  of  a  triangular 
area  with  respect  to  an  axis  parallel  to  the  base  and  containing  the 
centroid  ;  also  with  respect  to  the  base. 

Ans.  With  respect  to  central  axis,  /  =  b/i6/^,  k1  =  /i*/i8.  With 
respect  to  base,  /=  b/i*/ 12,  k1  =  /i2/6. 

6.  Prove  that  the  radius  of  gyration  of  an  elliptic  area  of  semi- 
axes  a  and  b,  with  respect  to  the  diameter  whose  length  is  2a,  is  b/2. 

404.  Product  of  Inertia  of  Plane  Area. —  The  sum  of  the  prod- 
ucts obtained  by  multiplying  every  element  of  a  plane  area  by  the 
product  of  its  distances* from  a  pair  of  rectangular  planes  is  called  the 
product  of  inertia  of  the  whole  area  with  respect  to  those  planes. 

The  most  important  case  is  that  in  which  the  two  planes  are  per- 
pendicular to  the  plane  of  the  given  area.      In  this  case  the  attention 


MOMENT    OF    INERTIA. 


341 


may  be  confined  to  the  lines  in  which  the  two  assumed  planes  intersect 
the  plane  of  the  given  area,  and  the  definition  may  be  stated  as  fol- 
lows : 

The  product  of  inertia  of  a  plane  area  with  respect  to  a  pair  of 
rectangular  axes  lying  in  its  plane  is  the  sum  of  the  products  obtained 
by  multiplying  every  element  of  area  by  the  product  of  its  distances 
from  the  two  axes. 

Reasoning  as  in  Art.  402,  it  is  seen  that  if  one  of  the  axes  is  an 
axis  of  symmetry  of  the  given  area,  the  product  of  inertia  is  equal  to 
zero.  It  will  be  shown  presently  that,  for  any  area  whatever,  there 
is  a  pair  of  axes  through  every  point  with  respect  to  which  the  prod- 
uct of  inertia  is  zero. 

405.  Products  of  Inertia  With  Respect  to  Different  Pairs  of 
Axes  Intersecting  in  the  Same  Point. —  Let  OX  and  OY(Fig.  164) 
be  any  pair  of  rectangular  axes  lying 
in  the  plane  of  a  given  area,  and  let 
B  represent  the  moment  of  inertia 
with  respect  to  OX,  A  the  moment 
of  inertia  with  respect  to  OY,  and  H 
the  product  of  inertia  with  respect  to 
OX  and  0  Y.  Take  a  second  pair  of 
rectangular  axes  OX',  OY',  the  an- 
gles XOX'  and  YOYf  being  each 
equal  to  6.  Let  H'  denote  the  product  of  inertia  with  respect  to 
this  new  pair  of  axes. 

If  x,  y  are  the  coordinates  of  any  point  referred  to  OX  and  0  Y, 
and  x ',  y'  its  coordinates  referred  to  OX'  and  OY' , 


Fig.  164. 


x'  =  x  cos  6  -J-  y  sin  6,     y'  = 
x'y'  =  — x2  sin  6  cos  6  -\-  xy(cos2  6 


x  sin  0  -\-  y  cos  6  ; 

sin2  6)  -\-  y2  sin  6  cos  6 


=  xy  cos  26  -j-  \{y2 — .r2)  sin  20. 

Hence         H'  =fx'y'dM 

=  cos  26  -fxydM  +  i  sin  2<9  •  [ffdM- 
=  H  cos  26  +  i(B  —  A)  sin  2(9. 


■fx2dM] 


If  the  moments  of  inertia  and  the  product  of  inertia  with  respect 
to  OX  and  0  Y  are  known,  H'  can  thus  be  computed  for  any  value 
oid. 


342  THEORETICAL    MECHANICS. 

406.  Product  of  Inertia  Zero  for  Certain  Axes. —  The  axes 
OX '  and  O  V  can  be  so  chosen  that  H'  is  zero.  For,  equating  the 
above  value  of  H'  to  zero  gives 

tan  2d  =  2HKA  —  B). 

This  equation  gives  a  real  value  of  0  for  any  real  values  of  A,  B 
and  H  ;  a  pair  of  axes  can  therefore  always  be  found  for  which  the 
product  of  inertia  is  zero.  Moreover,  there  is  but  one  pair  of  axes 
satisfying  this  condition  :  for  any  two  values  of  26  having  the  same 
tangent  differ  by  some  multiple  of  1800,  so  that  any  two  values  of  6 
satisfying  the  above  equation  differ  by  some  multiple  of  900. 

407.  Moments  of  Inertia  With  Respect  to  Different  Axes 
Through  the  Same  Point. —  The  moment  of  inertia  of  a  plane  area 
with  respect  to  each  of  two  rectangular  axes  through  a  given  point, 
and  the  product  of  inertia  with  respect  to  the  same  axes,  being 
known,  the  moment  of  inertia  with  respect  to  any  other  axis  through 
that  point  may  be  determined. 

Thus,  referring  to  Fig.  164,  let  I  be  the  moment  of  inertia  with 
respect  to  OX'.     Its  value  is 

I=fy'2dM=f(—xs'm  6  -\-y  cos  0)2dM 
=  sin2  6  -fx2dM  +  cos2  6  -ffdM  —  2  sin  6  cos  6  -fxydM 
=  A  sin2  6  +  B  cos2  6  —  2 H  sin  6  cos  6.  .         .         .     (1) 

If  OX  and  (9 Fare  the  pair  of  axes  for  which  the  product  of  in- 
ertia is  zero, 

I=A  sin2  0  +  B  cos2  0.      .  (2) 

Examples. 

1.  Find  the  moment  of  inertia  and  radius  of  gyration  of  a  rec- 
tangular area  of  sides  a  and  b  with  respect  to  a  central  axis  inclined 
300  to  the  side  a.  Ans.  k2  =  O2  +  3#2)/48. 

2.  Prove  that  the  moment  of  inertia  of  a  square  area  has  the  same 
value  for  all  axes  through  the  centroid. 

3.  Determine  the  moment  of  inertia  and  radius  of  gyration  of  a 
square,  the  length  of  whose  side  is  4  ins.,  with  respect  to  an  axis 
through  the  intersection  of  two  sides  and  inclined  at  an  angle  of  6o° 
to  one  side. 

[Take  the  axes  of  x  and  y  coincident  with  two  sides  of  the  square 
and  compute  A,  B,  H.  Since  His  not  zero,  equation  (1)  must  be 
used.  ] 


MOMENT    OF    INERTIA.  343 

4.  Show  that  the  sum  of  the  moments  of  inertia  of  a  plane  area 
with  respect  to  two  rectangular  axes  lying  in  its  plane  is  equal  to  the 
moment  of  inertia  with  respect  to  an  axis  perpendicular  to  the  plane 
of  the  area  and  containing  the  point  of  intersection  of  the  rectangular 
axes.  It  follows  that  this  sum  has  the  same  value  for  every  pair  of 
rectangular  axes  drawn  through  the  same  point. 

408.  Principal  Axes   and  Principal  Moments  of   Inertia. —  If 

the  values  of  the  moment  of  inertia  for  all  axes  through  a  given  point 
be  compared,  the  greatest  and  least  values  are  found  for  the  axes 
with  reference  to  which  the  product  of  inertia  is  zero. 

Thus,  the  value  of  /  given  by  equation  (2)  of  Art.  407  may  be 
expressed  in  the  form 

I=(A-B)sm*0  +  B.       ...     (3) 

Suppose  A  greater  than  B.  As  6  increases  from  o  to  90°,  the  value 
of /increases  from  B  to  A.  As  6  increases  from  900  to  1800,  the 
value  of  /  decreases  from  A  to  B.  Hence  A  is  the  greatest  and  B 
the  least  value  of  /. 

The  maximum  and  minimum  values  of  the  moment  of  inertia  for 
axes  through  a  given  point  are  called  principal  moments  of  inertia 
for  that  point.  The  corresponding  axes  are  called  principal  axes  for 
that  point. 

If  the  two  principal  moments  of  inertia  for  axes  passing  through 
a  given  point  are  equal,  the  moments  of  inertia  for  all  axes  through 
that  point  are  equal.     This  is  obvious  from  equation  (2). 

409.  Axis  of  Symmetry  a  Principal  Axis  at  Every  Point.  — 

If  an  area  possesses  an  axis  of  symmetry,  it  is  a  principal  axis  at  every 
point.  That  is,  the  axis  of  symmetry  and  any  line  perpendicular  to 
it  in  the  plane  of  the  area  are  principal  axes  at  their  point  of  inter- 
section. For  the  product  of  inertia  for  such  a  pair  of  axes  is  zero, 
as  may  be  proved  by  the  method  used  in  Art.  402. 

Examples. 

1 .  Determine  the  principal  moments  of  inertia  of  a  rectangle  of 
sides  4  ins.  and  6  ins.  for  axes  through  the  middle  point  of  the  longer 
side. 

2.  Determine  the  principal  moments  of  inertia  of  an  isosceles  tri- 
angle with  respect  to  axes  through  the  centroid. 

3.  Determine  the  principal  moments  of  inertia  of  an  area  con- 
sisting of  a  square  and  an  isosceles  triangle  whose  base  coincides  with 


344 


THEORETICAL    MECHANICS. 


a  side  of  the  square,  with  respect  to  axes  passing  through  the  vertex 
of  the  triangle. 

4.  Determine  the  principal  moments  of  inertia  of  the  area  de- 
scribed in  Ex.  3  with  respect  to  central  axes. 

5.  For  what  axis,  passing  through  the  intersection  of  two  sides 
of  a  rectangle,  is  the  moment  of  inertia  a  maximum? 

Taking  axes  of  x  and  y  coinciding  with  the  two  sides  of  the 
rectangle,  determine  the  values  of  A,  B  and  H  to  be  used  in  the 
formula  of  Art.  406  for  determining  0.  The  result  is  tan  26  = 
2>a6/2(a2  —  b2). 

6.  Determine  the  greatest  and  least  moments  of  inertia  of  a  rect- 
angle with  respect  to  axes  passing  through  the  intersection  of  two 
sides. 

410.  Radius  of  Gyration  With  Respect  to  Any  Axis  Through 
a  Given  Point.— Let  <9X  and  OY  (Fig.  165)  be  principal  axes  of  a 

given  plane  area,  A  being  the  moment 
of  inertia  with  respect  to  0  Y  and  B  the 
moment  of  inertia  with  respect  to  OX, 
a  and  b  the  corresponding  radii  of  gyra- 
tion, and  M  the  total  area,  so  that  A  = 
Ma2,  B  =  Mb2.     Let  /  be  the  moment 
of  inertia  and  k  the  radius  of  gyration 
with  respect  to  an  axis  OP,  inclined  to 
Then  (Art.  407) 
I  =  A  sin20  +  B  cos2d; 
.  • .     k2  =  a2  sin2 0  +  b2  cos2 6.  (4) 

411.  Inertia-Ellipse. —  If  in  Fig.  165  the  point  P  be  so  taken 
that  the  distance  OP  depends  in  some  assumed  manner  upon  the 
value  of  k,  the  point  P  will  describe  a  curve  as  6  varies.  If  this 
curve  is  known,  the  value  of  k  for  any  axis  can  be  determined.  The 
most  convenient  representation  results  from  the  assumption  that  OP 
is  inversely  proportional  to  k. 

Let  the  distance  OP  be  represented  by  r,  and  assume 
r  =  ab/k, 
r  being  thus  a  linear  magnitude.     From  equation  (4), 


Fig.  165. 
OX  at  an  angle  6. 


a2b2 


=  a2sW6  +  b2cos2d} 


(5) 


which  is  the  polar  equation  of  the  curve  described  by  the  point  P. 
\i  x  andjj/  are  the  rectangular  coordinates  of  P, 


MOMENT    OF    INERTIA.  345 

r  sin  6  =  y ;     r  cos  6  =  x ; 


and  the  rectangular  equation  of  the  curve  is 


-2 +-£=.,  •         •         •         •      (6) 

P  tf 


which  represents  an  ellipse  of  semi-axes  #  and  b. 

The  most  convenient  method  of  using  the  ellipse  depends  upon 
the  following  property  : 

If  a  tangent  be  drawn  to  the  ellipse,  inclined  at  angle  6  to  OX, 
the  perpendicular  distance  from  the  origin  to  this  tangent  (repre- 
sented by  p)  is  given  by  the  equation  * 

/  =  a2sm2d  +  b2  cos'd. 

Therefore,  comparing  with  equations  (4)  and  (5), 

p  =  abjr  =  k. 

That  is,  the  radius  of  gyration  with  respect  to  any  axis  through  O  is 
equal  to  the  perpendicular  distance  between  this  axis  and  the  parallel 
tangent  to  the  ellipse. 

This  ellipse  is  called  the  ellipse  of  inertia  for  the  point  O.  It  is 
seen  that  its  principal  diameters  lie  in  the  principal  axes  of  inertia 
(Art.  408) ;  the  principal  semi-axis  lying  in  OX  is  the  radius  of  gyra- 
tion with  respect  to  OY;  and  the  principal  semi-axis  lying  in  OY  is 
the  radius  of  gyration  with  respect  to  OX. 

Central  ellipse. —  For  every  point  in  the  plane  of  the  area,  there 
is  a  definite  inertia-ellipse  whose  center  is  at  that  point.  The  ellipse 
whose  center  is  at  the  centroid  of  the  area  is  called  the  central 
ellipse. 

Examples. 

1.  Determine  the  central  ellipse  for  a  rectangular  area. 

2.  Show  that  the  central  ellipse  for  the  area  of  an  ellipse  is  a  sim- 
ilar ellipse.     What  are  its  semi-axes?  Ans.  a/ 2  and  6/2. 

3.  Determine  the  central  ellipse  for  an  isosceles  triangle. 

4.  Show  that  the  central  ellipse  for  an  equilateral  triangle  is  a 
circle,  and  determine  its  radius. 

Ans.   (#i/6)/i2,  if  a  is  the  side  of  the  triangle. 

5.  Determine  the  central  ellipse  for  a  regular  hexagon.  

Ans.  A  circle  of  radius  equal  to  tfl/5/24. 

*  Smith's  "Conic  Sections,"  Art.  115. 


CHAPTER   XX. 

MOTION    OF    A    RIGID     BODY  :     TRANSLATION  J     ROTATION     ABOUT   A 

FIXED    AXIS. 

§  i.   Simple  Motions  of  a  Rigid  Body. 

412.  Coordinates  of  Position — Any  quantities  whose  values 
serve  to  specify  the  position  of  every  particle  of  a  body  are  called  its 
coordinates  of  position. 

It  may  be  shown  that  six  coordinates  are  sufficient  to  specify  the 
position  of  a  rigid  body  which  is  free  to  move  in  three  dimensions  of 
space.     These  six  quantities  may  be  chosen  in  various  ways. 

To  show  that  six  coordinates  are  sufficient,  notice  that  if  three 
points  in  the  body,  not  lying  in  a  right  line,  are  fixed,  every  point  is 
fixed.  Let  A,  B,  Cbe  three  such  points.  The  position  of  any  one, 
as  A,  is  given  by  its  three  rectangular  coordinates  xx,  yx,  zx  with  re- 
spect to  an  assumed  set  of  fixed  axes.  The  position  of  B  is  then 
determined  by  the  values  of  any  two  of  its  three  rectangular  coor- 
dinates x.z ,  y% ,  z% ,  since  its  distance  from  A  is  fixed.  When  the  posi- 
tions of  A  and  B  are  given,  that  of  C  is  determined  by  the  value  of 
one  of  its  three  rectangular  coordinates,  since  its  distances  from  A 
and  B  are  fixed.  Thus,  to  completely  specify  the  positions  of  A,  B 
and  C  (and  therefore  of  all  points  of  the  body)  it  is  sufficient  to 
assign  values   to  six  of   the   nine   quantities  xx ,  ylt  zlf  x2 ,  y2 ,  z2 , 

413.  Degrees  of  Freedom. —  The  six  coordinates  whose  values 
specify  the  position  of  a  rigid  body  may  be  varied  independently  of 
one  another ;  for  example,  one  may  be  varied  while  the  other  five 
remain  constant.  Such  a  variation  causes  the  body  to  move  in  a 
particular  way.  This  is  often  expressed  by  the  statement  that  the 
body  possesses  one  degree  of  freedom  for  every  coordinate.  A  rigid 
body  whose  motion  is  unrestricted  possesses  six  degrees  of  freedom. 

By  imposing  restrictions  on  the  motion  of  a  body  the  number  of 
degrees  of  freedom  may  be  artificially  reduced.  The  following  dis- 
cussion will  deal  mainly  with  motions  so  restricted  as  greatly  to 
simplify  the  application  of  dynamical  principles. 


SIMPLE    MOTIONS    OF    A    RIGID    BODY.  347 

414.  Motion  of  Translation. —  If  the  velocities  of  all  particles  of 
a  body  are  at  every  instant  equal  in  magnitude  and  direction,  the 
body  is  said  to  have  a  motion  of  tra?islation. 

In  a  motion  of  translation  a  particle  of  the  body  may  describe 
any  path  in  space.  If  no  additional  restriction  is  imposed,  the  body 
possesses  three  degrees  of  freedom.  The  coordinates  of  position  of 
the  body  may,  for  example,  be  the  rectangular  coordinates  of  any  one 
particle.     The  paths  of  all  particles  are  alike  in  all  respects. 

415.  Rotation  About  a  Fixed  Axis.— If  all  particles  of  a  rigid 
body  describe  circles  whose  centers  lie  in  a  certain  fixed  line,  the 
motion  is  a  rotation.  The  fixed  line  containing  the  centers  of  the 
circles  described  by  the  particles  is  the  axis  of  rotation. 

If  the  motion  is  restricted  to  a  rotation  about  a  fixed  axis,  the 
body  possesses  only  one  degree  of  freedom.  The  position  of  the 
body  is  completely  specified  by  the  value  of  one  coordinate,  which 
may,  for  example,  be  the  angle  between  two  planes  containing  the 
axis  of  rotation,  one  of  which  is  fixed  in  space  and  the  other  fixed  in 
the  body. 

Let  Fig.  166  represent  a  section  of  the  body  by  a  plane  perpen- 
dicular to  the  axis  of  rotation,  and  let  OX  and  OA  be  the  traces  of 
planes  containing  the  axis,  OX  being 
fixed  while  OA  turns  with  the  body. 
The  position  of  the  body  at  any  instant 
is  completely  specified  by  the  value  of  the 
angle  A  OX.  Call  this  6.  If  0  is  known 
as  a  function  of  the  time  the  motion  is 
completely  determined. 

Angular  motion. —  All   planes  con- 
taining the  axis  of  rotation  or  parallel  to  Fig.  166.    . 
it  turn  through  equal  angles  during  any 

definite  time  ;  hence  the  angular  motion  of  any  one  of  them  as  OA 
may  be  called  the  angular  motion  of  the  body.  This  angular  motion 
may  be  specified  as  in  Art.  281. 

The  angle  turned  through  in  any  definite  time  is  the  angidar 
displacement.  The  angular  displacement  per  unit  time  is  the  angular 
velocity.  The  increment  of  the  angular  velocity  per  unit  time  is  the 
angular  acceleration. 

The  value  of  the  angular  displacement  of  the  body  during  any 
time  At  is 


348  THEORETICAL    MECHANICS. 

if  0l  and  02  are  the  initial  and  final  values  of  6. 

If  ft)  represents  the  angular  velocity  of  the  body  and  <j>  its  angular 
acceleration  at  any  instant,  their  values  may  be  expressed  as  in  Art. 

ft)  ==  d0/dt;     (j>  =  dco/dt  ==  d20/dt\ 

416.  Plane  Motion. —  Plane  motion  of  a  rigid  body  is  a  motion 
in  which  all   particles  move  in  parallel  planes.     This  is  also  called 

uniplanar  motion. 

A  body  whose  motion  is  thus  restrict- 
ed has  three  degrees  of  freedom.  This 
is  evident  by  considering  that  the  posi- 
tion may  be  completely  specified  by  the 
values  of  three  coordinates,  as  follows: 

Let  OX,  OY  (Fig.  167)  be  a  pair  of 
fixed  rectangular  axes  lying  in  a  plane 
q  ~X     parallel  to  the  motion.     Let  A  and  B  be 

pIG#  !67-  the  positions  of  any  two  particles  of  the 

body  lying  in  the  plane  XO  V ;  and  let 
x,  y  denote  the  rectangular  coordinates  of  A  and  0  the  angle  be- 
tween AB  and  OX.  The  position  of  the  body  is  completely  specified 
if  values  are  assigned  to  x,  y  and  6. 

The  general  case  of  plane  motion  will  be  considered  in  Chapter 
XXI. 

417.  Determination  of  Motion  of  Mass-Center. —  It  has  been 
shown  (Art.  380)  that  the  motion  of  the  center  of  mass  of  any  system 
of  particles  depends  only  upon  the  external  forces.  To  determine 
this  motion,  the  entire  mass  of  the  system  is  conceived  to  be  concen- 
trated at  the  center  of  mass  and  to  be  acted  upon  by  forces  which  at 
every  instant  are  equal  in  magnitude  and  direction  to  the  external 
forces  acting  upon  the  particles  of  the  system. 

As  an  example  of  the  application  of  this  principle,  consider  the 
motion  of  a  projectile.  If  a  body  is  thrown  in  any  way  and  is  then 
acted  upon  by  no  force  except  its  weight,  the  motion  of  the  center  of 
mass  may  be  determined  by  the  results  of  Art.  295,  which  were  de- 
duced for  the  case  of  a  particle.  For  the  resultant  of  the  external 
forces  is  constant  in  magnitude  and  direction  during  the  motion. 
The  same  will  be  true  of  two  or  more  bodies  connected  by  cords  or 
otherwise ;  or  of  any  set  of  bodies  whatever,  regarded  as  a  system. 


ROTATION    ABOUT    A    FIXED    AXIS.  349 

418.  Condition  That  Motion  May  Be  a  Translation If  the 

motion  of  a  rigid  system  throughout  any  interval  is  known  to  be  a 
translation,  the  equations  of  motion  of  the  mass-center  suffice  to 
determine  .the  motion  of  every  particle,  since  the  paths  of  all  particles 
are  alike  in  every  respect.  It  remains  to  consider  under  what  con- 
ditions the  motion  will  be  a  translation.  This  question  may  be 
answered  by  applying  D'Alembert's  principle  (Art.  390)  that  the 
external  forces  and  the  effective  forces  form  equivalent  systems* 

Resultant  effective  force. — In  a  motion  of  translation,  the  accel- 
erations of  all  particles  of  the  body  are  at  every  instant  equal  in 
magnitude  and  direction.  The  effective  forces  (each  being  equal  to 
the  product  of  the  mass  of  a  particle  into  its  acceleration)  form  a 
system  of  parallel  forces  whose  magnitudes  are  proportional  to  the 
masses  of  the  particles.  The  resultant  of  this  system  evidently  acts 
in  a  line  passing  through  the  mass-center,  and  its  value  is  Mp  if  M 
is  the  total  mass  and/  the  acceleration. 

Since  the  resultant  external  force  and  the  resultant  effective  force 
are  at  every  instant  equal  in  all  respects  (magnitude,  direction  and 
line  of  action),  it  follows  that  the  line  of  action  of  the  resultant  of  the 
external  forces  must  pass  through  the  mass-center  of  the  system  in 
order  that  the  motion  may  continue  translatory. 

The  converse  of  this  proposition  is  not  necessarily  true.  But  if, 
at  any  instant,  all  particles  have  equal  and  parallel  velocities,  the 
motion  will  continue  to  be  translatory  if  the  resultant  of  the  external 
forces  acts  in  a  line  containing  the  center  of  mass. 

§  2.  Rotation  About  a  Fixed  Axis. 

419.  Rotation  Under  Any  Forces. —  If  a  body  is  constrained  to 
rotate  about  a  fixed  axis,  it  possesses  but  one  degree  of  freedom,  and 
one  dynamical  equation  is  sufficient  to  determine  the  motion  if  all 
external  forces  are  known.  Generally,  however,  certain  of  the  ex- 
ternal forces  are  unknown,  and  additional  equations  are  needed  in 
order  to  determine  them. 

The  method  by  which,  practically,  the  motion  of  a  body  can  be 
restricted  to  rotation  about  a  fixed  axis,  is  by  means  of  a  hinge  joint 
(Art.  42)  or  something  equivalent.     The  force  or  forces  exerted  upon 

*  The  reader  should  bear  in  mind  the  remarks  made  in  Art.  387  as  to  the 
meaning  of  this  principle. 


350  THEORETICAL    MECHANICS. 

the  body  by  the  hinge  are  in  general  unknown,  and  are  to  be  deter- 
mined by  means  of  the  dynamical  equations.  D' Alembert's  principle 
yields  three  independent  equations  for  plane  motion  (Art.  388),  and 
these  suffice  to  determine  the  motion  and  the  restraining-  forces  in 
case  of  rotation  about  a  fixed  axis.*  At  least  one  of  the  three  equa- 
tions must  be  an  equation  of  moments.  For  the  other  two,  equations 
of  resolution  will  usually  be  preferred. 

420.  Moments  of  Effective  Forces  About  Axis  of  Rotation. — 

Let  D' Alembert's  principle  be  applied,  taking  moments  about  the 
fixed  axis  of  rotation.  The  sum  of  the  moments  of  the  effective  forces 
about  this  axis  may  be  computed  as  follows  : 

Effective  force  for  a  particle. — Every  particle  of  a  body  rotating 
about  a  fixed  axis  describes  a  circle  with  center  in  that  axis.  For 
any  particle  the  resultant  acceleration  may  be  specified  by  its  com- 
ponents along  the  tangent  and  normal  to  the  circle.  If  r  is  the 
distance  of  the  particle  from  the  center  of  rotation  and  v  its  velocity, 
the  two  components  are  as  follows  (Art.  284): 
Tangential  component,  dvjdt\ 

Normal  component,  v2jr}  directed  toward  the  center. 
If  co  is  the  angular  velocity  and  <f>  the  angular  acceleration,  v  is 
equal  to  rco,  and  since  r  is  constant, 

dv\dt  =  r{dco\df)  =  rep ; 
v2/r  ==  r2co2/r  ==  rco2. 
If  the  mass  of  the  particle  is  m,  the  effective  force  is  equivalent  to 
the  two  components 

mr<j>,      directed  along  the  tangent  to  the  circle ; 
mrco\     directed  toward  the  center. 

*  If  the  applied  forces  tend  to  give  the  body  motion  parallel  to  the  axis  of 
rotation,  the  method  of  constraint  must  be  such  as  to  counterbalance  this 
tendency,  and  the  restraining  forces  will  have  components  parallel  to  the  axis. 
Again,  the  inertia  of  the  particles  will  in  general  tend  to  cause  a  departure 
from  the  prescribed  plane  motion,  which  tendency  must  be  resisted  by  the 
constraining  forces.  In  the  most  general  case,  six  dynamical  equations  are 
needed  in  order  to  completely  determine  the  constraining  forces.  The  fol- 
lowing discussion  is  restricted  to  the  case  in  which  the  applied  forces  and  the 
distribution  oi  mass  are  such  that  there  is  no  tendency  to  depart  from  a 
condition  of  plane  motion,  so  that  the  three  dynamical  equations  for  plane 
motion  are  sufficient  to  determine  the  constraining  forces  as  well  as  the 
motion. 


ROTATION    ABOUT    A    FIXED  AXIS.  35 1 

Since  the  moment  of  the  latter  component  about  the  axis  of  rotation 
is  zero,  the  moment  of  the  effective  force  is  equal  to 

mr2<j>. 

Total  moment  of  effective  forces. — If  the  masses  of  the  particles  of 
the  system  are  mx>  m%i  .  .  .  ,  and  their  distances  from  the  axis 
of  rotation  rx ,  rt ,  .  .  .  ,  the  sum  of  the  moments  of  the  effective 
forces  has  the  value 

(m,r2  -f-  m2r2  +     .     .     .     )<£  =  I(j>, 
I  being  the  moment  of  inertia  of  the  system  with  respect  to  the  axis 
of  rotation. 

421.  Equation  of  Angular  Motion. —  The  equation  of  angular 
motion  is  obtained  by  equating  the  sum  of  the  moments  of  the  ef- 
fective forces  to  the  sum  of  the  moments  of  the  external  forces.  Let 
L  denote  the  algebraic  sum  of  the  moments  of  the  external  forces 
with  respect  to  the  axis  of  rotation  ;  then 

L  =  I<f>. 

If  6  denotes  the  angle  between  some  line  in  the  system  and  a 
fixed  reference  line,  the  second  member  of  the  equation  may  be  writ- 
ten in  either  of  three  forms,  as  follows  : 

L  =  f(t>  =  I(dco/dt)   =  /(d'^/dt2).  .  .        (I) 

If  L  is  known,  or  can  be  expressed  in  terms  of  one  or  both  of 
the  variables  0  and  /,  the  integration  of  equation  (i)  serves  to  deter- 
mine the  motion  completely,  provided  sufficient  initial  conditions  are 
known  for  the  determination  of  the  constants  of  integration. 

422.  Hinge  Reaction. —  The  rotation  of  a  system  about  a  fixed 
point  is  a  case  of  constrained  motion 
(Art.  303);  a  certain  condition  is  im- 
posed upon  the  motion  without  specifi- 
cation of  the  forces  necessary  to  maintain 
that  condition.  In  the  case  supposed, 
the  constraint  must  be  effected  by  some- 
thing equivalent  to  a  hinge  at  the  center 
of  rotation.  Thus,  if  O  (Fig.  168)  is 
the  center  of  rotation,  any  tendency  of 
the  body  to  depart  from  the  stated  mo-  Fig.  168. 

tion  is  resisted  by  forces  equivalent  to  a 

single  force  applied  at  O.    This  force  may  be  called  the  hinge  reaction. 


352 


THEORETICAL    MECHANICS. 


Suppose  the  hinge  to  consist  of  a  fixed  cylindrical  pin  fitting  into 
a  hole  in  the  rotating  body.  If  the  surfaces  were  smooth  at  the  point 
of  contact,  the  pressure  of  the  pin  upon  the  body  would  act  in  a  line 
passing  through  the  axis  of  rotation.  In  reality  it  will  depart  from 
this  direction  because  of  the  friction.  Let  F  be  the  frictional  force 
and  R  the  normal  force ;  then  the  moment  of  R  with  respect  to  the 
axis  of  rotation  is  zero,  so  that  R  does  not  affect  the  value  of  L. 
The  moment  of  F  must  be  included  in  L,  its  sign  being  always  oppo- 
site to  that  of  the  existing  rotation  of  the  body.  In  the  following 
discussion  the  friction  will  either  be  neglected  or  will  be  assumed  to 
be  included  among  the  known  external  forces  whose  moment  is  L. 
The  hinge-reaction  will  be  regarded  as  acting  always  in  a  line  which 
intersects  the  axis  of  rotation. 

423.  Equations  of  Motion  of  Mass-Center. —  The  value  of  the 
hinge-reaction  may  be  determined  by  writing  the  equations  of  motion 
of  the  center  of  mass.  If  P  is  the  vector  sum  of  all  external  forces 
(including  the  hinge-reaction),  M  the  mass  of  the  body,  and/  the 

acceleration  of  the  mass-center,  these 
quantities  satisfy  the  vector  equation 
P  =  Mp.  This  has  been  proved 
(Art.  380)  for  any  system  of  particles, 
whether  rigid  or  not. 

In  the  case  of  a  body  rotating 
about  a  fixed  axis,  p  may  be  ex- 
pressed in  terms  of  the  angular  veloc- 
•  ity,  angular  acceleration,  and  distance 
of  the  center  of  mass  from  the  axis  of  rotation.  Let  Fig.  169  repre- 
sent a  section  of  the  body  by  a  plane  perpendicular  to  the  axis  of 
rotation,  containing  the  mass-center  G ;  OG  =  a  =  perpendicular 
from  mass-center  on  axis  of  rotation  ;  0  =  angle  between  OG  and  a 
fixed  plane  OA.     Then/  is  equivalent  to  the  two  components 

a(d26/dt2)  perpendicular  to  OG,     a(d6/dt)2  in  direction  GO. 

Let  all  external  forces  be  resolved  into  components  respectively 
perpendicular  and  parallel  to  GO.  Let  T,  N  be  the  components  of 
the  hinge-reaction ;  Pt ,  P„  the  components  of  all  other  external 
forces.  The  positive  directions  for  the  resolution  must  agree  with 
those  used  in  resolving/  ;  they  are  shown  by  the  arrows  in  Fig.  169. 
The  figure  shows  P„  and  Pt  as  if  applied  at  G,  but  it  is  not  to  be 


Fig.  169. 


ROTATION    ABOUT    A    FIXED    AXIS.  353 

inferred  that  this  is  necessarily  the  fact ;  we  are  not  here  concerned 
with  their  lines  of  action. 

The  equations  of  motion  of  the  mass-center  may  then  take  the 
forms 

Pt+  T=  Ma(d26/dt2);        .         .         .     (2) 
Pn  +  N=  Ma(deidt)\         .         .         .     (3) 

424.  Complete  Solution  of  Problem  of  Rotation. — The  com- 
plete solution  of  the  problem  is  contained  in  equations  (1),  (2)  and 
(3),  which  for  convenience  will  be  written  together  here : 

L  =  I(d-16ldt2);  .  .  .     (1) 

Pt  +  T  =  Ma(d*ejdt*)\       .        .        .     (2) 
Pn  + N  =  Ma{dd\df)\  .         .         .     (3) 

Equation  (1)  does  not  involve  the  hinge-reaction,  and  if  all  other 
external  forces  are  known,  the  motion  is  determined  by  integrating 
this  equation.  Equations  (2)  and  (3)  serve  to  determine  T  and  iV, 
the  components  of  the  hinge-reaction,  after  the  motion  is  known. 
It  will  be  observed  that  the  complete  integration  of  (1)  is  not  neces- 
sary in  order  that  T  and  N  may  be  determined  from  (2)  and  (3). 

Examples. 

1.  A  body  of  mass  M  rotates  about  a  fixed  axis  distant  a  from  the 
mass-center.  If  no  force  acts  upon  the  body  except  the  hinge-reac- 
tion, and  if  its  angular  velocity  at  a  certain  instant  is  col ,  determine 
(a)  the  subsequent  motion  and  (&)  the  hinge  reaction. 

(a)  In  equation  (1),  L  =  o,  hence 

d'djdt2  =  dcaldt  =  o ; 
ddldt  =  to  =  constant  =  co1 ; 
0  =  ay  J  -f  $1  ; 

6l  being  the  value  of  6  when  t  =  o.     The  angular  velocity  thus  re- 
mains constant. 

(b)  The  resultant  acceleration  of  the  mass-center  is  aco\  directed 
toward  the  center  of  rotation  ;  the  resultant  effective  force  is  there- 
fore Maco2  in  that  direction.  Since  the  hinge-reaction  R  is  the  only 
external  force,  it  must  be  equal  in  all  respects  to  the  resultant  effective 
force. 

It  is  evident  that  this  result  is  given  by  equations  (2)  and  (3). 
Since  Pt  and  Pn  are  both  zero,  (2)  becomes  T  =  0,  and  (3)  reduces 
to  N  =  Maw2. 


354  THEORETICAL    MECHANICS. 

2.  A  body  of  mass  ioo  lbs.  rotates  about  a  smooth  hinge  2  ft. 
from  the  mass-center,  being  acted  upon  by  no  external  force  except 
the  hinge-reaction.  If  at  a  certain  instant  it  is  rotating  at  the  rate  of 
2  rev. -per-sec. ,  what  is  the  subsequent  motion,  and  what  is  the  value 
of  the  hinge-reaction  ? 

Ans.   Hinge-reaction  =  N  =  3,200  ir2/g  pounds- weight. 

3.  A  body  of  mass  200  lbs. ,  acted  upon  by  gravity,  rotates  about 
a  smooth  horizontal  axis  passing  through  the  mass-center.  If  at  a 
certain  instant  the  rate  of  rotation  is  3  rev. -per-sec. ,  determine  the 
subsequent  motion  and  the  hinge-reaction. 

[Notice  that  if  the  axis  of  rotation  contains  the  mass-center, 
a  =  o,  and  equations  (2)  and  (3)  simplify.] 

4.  A  homogeneous  cylinder  of  mass  200  lbs.  and  diameter  2  ft. 
rotates  about  its  axis  of  figure  (horizontal)  under  the  action  of  a  con- 
stant pull  of  5  lbs.  applied  to  the  free  end  of  a  string  which  is  wrapped 
around  the  cylinder.  At  a  certain  instant  it  is  rotating  at  the  rate  of 
200  rev.-per-min.  against  the  pull.  Determine  the  subsequent  mo- 
tion. When  will  the  body  come  to  rest  ?  Determine  the  hinge- 
reaction,  assuming  the  pull  to  be  vertically  downward. 

Ans.   It  will  come  to  rest  in  13  sec.    Hinge-reaction  =  205  lbs. 

5.  In  the  preceding  example,  let  the  tension  in  the  string  be  due 
to  a  suspended  weight  of  5  lbs. ,  the  remaining  data  being  as  before. 
Solve  the  problem  completely. 

[Write  the  equation  of  angular  motion  of  the  cylinder  and  the 
equation  of  linear  motion  of  the  suspended  weight ;  then  eliminate 
the  tension  in  the  string.  ] 

Ans.  It  will  come  to  rest  in  13.7  sec.  Hinge-reaction  = 
204.1  lbs. 

6.  A  wheel-and-axle  of  total  mass  60  lbs.  is  set  rotating  by  a  con- 
stant tension  of  7  lbs. -force  in  the  rope  which  unwinds  from  the  axle. 
The  radius  of  gyration  of  the  body  with  respect  to  the  axis  of  rotation 
is  10  ins.  and  the  radius  of  the  axle  is  6  ins.  Required  (a)  the  angular 
velocity  after  2  sec. ,  (d)  the  angle  turned  through  in  2  sec. ,  and  (Y)  the 
pressure  on  the  hinge.     (Assume  no  friction. ) 

Ans.   (a)  o.  1 68^  rad.  -per-sec.    (fi)  o.  1 68g*  rad.    (c)  6jg  poundals. 

7.  In  the  preceding  example,  instead  of  a  tension  of  7  lbs.,  as- 
sume a  tension  due  to  the  weight  of  a  suspended  body  of  7  lbs. 
mass  ;  solve  the  problem  completely. 

8.  In  example  6,  what  weight  suspended  from  the  rope  would 
produce  a  tension  of  7  lbs. -force?  Ans.   7.31  lbs. 

425.  Compound  Pendulum. —  A  rigid  body  rotating  about  a 
horizontal  axis  under  the  action  of  no  external  force  except  gravity 
and  the  hinge  reaction,  is  called  a  compound  pendulum. 

In  Fig.  170,  let  the  axis  of  rotation,  perpendicular  to  the  plane  of 


ROTATION    ABOUT    A    FIXED    AXIS. 


355 


the  figure,  be  projected  at  O,  and  let  A  be  the  center  of  mass.      Let 
B  be  the  lowest  point  reached  by  A,  and  A0  the  highest  point.     Let 
angle  AOB  =  0,  A0OB  =  0O,  AO  =  tf,  M=  mass  of  body,  k  = 
its  radius  of  gyration  with  respect  to  the 
axis  of  rotation,  /  =  Mk2  =  moment  of  in- 
ertia with  respect  to  that  axis. 

The  only  external  force,  except  the 
hinge  reaction,  is  the  weight  of  the  body, 
its  value  in  kinetic  units  being  Mg  and  its 
direction  vertically  downward.  The  mo- 
ment of  this  force  about  the  axis  of  rota- 
tion is 

L  —  —  Mga  sin  0, 

the  negative  sign  being  used    because,  in 

writing  the  equation  of  angular  motion,  the 

positive  direction  for  L  must  agree  with  that  for  0.   Equation  (i) 

therefore  reduces  to  the  form 

d20/dt2  =  —(ga/&)  sin  0.  (4) 

The  integration  of  this  equation  will  determine  the  motion. 

Equivalent  simple  pendulum.  —  The  equation  of  motion  of  a 
simple  pendulum  (a  particle  suspended  freely  from  a  fixed  point  by 
a  flexible  string  without  weight)  was  found  in  Art.  309  to  be 

d20/dt2  =  —  (g/l)  sin  0, 

I  being  the  distance  of  the  particle  from  the  point  of  suspension. 
This  is  identical  with  the  equation  just  derived  for  the  compound 
pendulum  if 

/  =  &/a. 

The  motion  of  the  compound  pendulum  is  therefore  the  same  as 
that  of  a  simple  pendulum  of  length  k2\a. 

Let  OA  (Fig.  170)  be  produced  to  a  point  0'  so  taken  that 
00'  =  I  =  klja.  The  point  O  is  called  the  center  of  suspension 
and  O'  the  center  of  oscillation.  The  following  considerations  show 
that  as  O  approaches  the  mass-center  O'  recedes  from  it. 

Let  k9  be  the  radius  of  gyration  with  respect  to  an  axis  through 
the  mass-center  A  parallel  to  the  axis  of  rotation  ;  then 

k2  =  k2  +  a2 ;     /  =  P/a  =  a  +  k2ja ; 
,\      (/  —  d)a  =  k2,     or     OA  X  0'A=  k*. 


356  THEORETICAL    MECHANICS. 

That  is,  the  product  of  the  distances  of  0  and  0 '  from  the  mass- 
center  is  constant. 

If  the  center  of  suspension  0  is  very  near  the  mass-center,  the 
value  of  /  becomes  very  great ;  if  0  is  very  distant  from  the  mass- 
center,  /  differs  little  from  a. 

From  the  above  relation  between  OA  and  O'A  it  is  seen  that  if 
O'  be  made  the  center  of  suspension,  O  becomes  the  center  of  oscil- 
lation. 

For  the  method  of  integrating  equation  (4),  and  of  determining 
the  time  of  a  small  oscillation,  reference  may  be  made  to  Art.  309. 
All  the  results  there  found  for  a  simple  pendulum  may  obviously  be 
applied  to  an  equivalent  compound  pendulum. 

The  first  integration  of  equation  (4)  gives 

(d6jdt)2  =  (2£tf//P)(C0S  0  —  cos  0O).      .  .      (5) 

The  solution  of  this  equation  involves  an  elliptic  integral.  But  for 
the  case  of  a  pendulum  oscillating  through  a  small  angle,  an  ap- 
proximate solution  is  obtained  by  putting  6  for  sin  6.  With  this 
substitution,  the  integration  of  equation  (4),  subject  to  the  condi- 
ditions  d0/dt  =  o  when  6  =  00 ,  6  =  o  when  /  =  o,  gives 

6  =  e0  sin  (/  \/~ga\¥).  .  .         .     (6) 

Examples. 

1.  If  the  center  of  suspension  of  a  homogeneous  circular  disc 
of  uniform  thickness  is  in  the  circumference,  determine  the  center 
of  oscillation  (a)  if  the  axis  of  suspension  lies  in  the  plane  of  the  disc 
and  (b)  if  it  is  perpendicular  to  that  plane. 

Ans.   (a)  00'  =  5^/4;  (b)  00'  =  30/2  ;  a  being  radius  of  disc. 

2.  The  center  of  suspension  of  a  homogeneous  rectangular  plate 
of  sides  a  and  b  is  distant  r  from  the  center  of  mass  ;  required  the 
center  of  oscillation,  the  axis  being  perpendicular  to  the  plane  of  the 
rectangle.     Ans.   {a1  +  ^2)/i2r  =  distance  from  center  of  mass. 

3.  If,  in  the  preceding  example,  a  =  40  ins. ,  b  =  2  ins.  and 
r=  18  ins.,  determine  the  center  of  oscillation  and  the  time  of  a 
small  oscillation.      Ans.  Time  of  oscillation  =  0.81  sec,  nearly. 

4.  A  homogeneous  straight  bar,  whose  cross-section  is  very  small 
in  comparison  with  its  length,  is  suspended  at  any  point.  Determine 
the  position  of  the  center  of  oscillation. 

5.  A  homogeneous  bar  of  small  uniform  cross -section,  6  ft.  long, 
is  suspended  so  as  to  rotate  without  retardation  by  friction.  Deter- 
mine the  length  of  the  equivalent  simple  pendulum  when  the  distance 


ROTATION   ABOUT    A    FIXED    AXIS.  357 

of  the  point  of  suspension  from  the  end  has  each  of  the  following 
values  :  o,  i  ft.,  2  ft.,  3  ft.      For  what  position  of  the  point  of  sus- 
pension will  the  length  of  the  equivalent  simple  pendulum  be  20  ft? 
Ans.  In  first  and  third  cases,  /  =  4  ft. 

6.  Determine  the  hinge-reactions  in  case  of  a  compound  pendu- 
lum. 

Substitute  in  equations  (2)  and  (3)  (Art.  424)  and  solve  for  T 
and  N.  The  only  external  force  aside  from  the  hinge-reaction  is  the 
weight  of  the  body  ;  its  components  are 

Pt  =  —Mg  sin  6,     Pn  ==  —Mg  cos  6. 

Equation  (4)  gives  the  value  of  d20/dt2,  and  (5)  the  value  of  {dO/dt)2. 
Hence 

T  =  Mg(i  —a2 jk1)  sin  0 ;       .  .         .     (7) 

N  =  Mg[(2a2/&2)<icos  6  —  cos  0O)  +  cos  0].         .     (8) 

7.  With  data  as  in  case  (a)  of  Ex.  1,  let  00  =  900.  Determine 
the  magnitude  and  direction  of  the  hinge-reaction  when  0  =  o,  900 
and  450  respectively. 

Ans.  When  6  =  o,  T  =  o,  N  ==  13^/5. 

8.  In  what  position  of  the  body  has  the  angular  velocity  its  great- 
est value  ?  Show  that  if  this  greatest  value  exceeds  a  certain  limit 
6Q  is  imaginary. 

9.  Get  a  first  integral  of  equation  (4),  determining  the  constant  by 
the  condition  that  dO/dt  =  co'  when  0  =  o. 

Ans.  co2  =  {dGjdff  =  co'2  —  {2ga\k2){\  —  cos  0). 

426.  Resultant  Effective  Force. —  The  resultant  of  a  system  of 
coplanar  forces  is  a  single  force  or  a  couple.  If  a  single  force,  its 
magnitude  and  direction  are  determined  if  its  resolved  parts  in  two 
directions  are  known  ;  if,  in  addition,  its  moment  about  any  point  be 
known,  the  line  of  action  is  determined. 

In  case  of  a  body  rotating  about  a  fixed  axis,  the  resultant  effective 
force  is  easily  determined  when  the  angular  velocity  and  angular 
acceleration  are  known.  If  M  is  the  mass,  /  =  Mk1  the  moment 
of  inertia  with  respect  to  the  axis  of  rotation,  p  the  acceleration  of 
the  mass-center,  co  =  dO/dt  the  angular  velocity,  and  cf>  =  dco/dt 
=  d26jdt2  the  angular  acceleration,  the  value  of  the  resultant  effective 
force  is  as  follows  : 

Its  magnitude  is  Mp ;  its  direction  is  that  of  p ;  its  moment  with 
respect  to  the  axis  of  rotation  (Art.  420)  is  Mk2<j>,  hence  its  line  of 
action  is  at  a  distance  k2<j>/p  from  the  axis  of  rotation.  This  result 
may  conveniently  be  put  in  another  form,  as  follows : 


358  THEORETICAL    MECHANICS. 

In  Fig.  171,  let  O  represent  the  axis  of  rotation,  A  the  mass- 
center,  and  O'  the  point  in  which  the  line  of  action  of  the  resultant 
effective  force  Mp  intersects  OA.  Let  OA  =  a,  00'  =  h.  Let 
Mp  be  replaced  by  two  components  acting  at  Or,  one  acting  in  the 

direction  O'O,  the  other  perpendicular 
to  that  direction.  The  components  of 
p  in  these  directions  are  axo2  and  a<f> ; 
hence  the  components  of  Mp  are 

Maco*     and     Ma<f>. 

The  sum  of  the  moments  of  these  com- 
ponents about  O  is 

Ma/16, 
Fig.  171. 

which    must    equal  the  moment    of  the 

resultant  effective  force.     But  (Art.  420)  this  is  also  equal  to 

Mkl$ ; 

hence  ah  =  £2,     or     h  =  k2/a. 

The  point  O'  thus  coincides  with  the  point  called  the  center  of 
oscillation  in  case  of  a  compound  pendulum. 

Resultant  couple. — If  the  axis  of  rotation  contains  the  center  of 
mass,/  =  o,  hence  the  vector  sum  of  the  effective  forces  is  zero. 
Their  moment  Mk2<f>  is  not  in  general  zero.  In  this  case  the  re- 
sultant effective  force  is  a  couple. 

Resultant  zero. — If  the  axis  of  rotation  contains  the  mass-center 
and  the  angular  velocity  is  constant,  the  ipoment  of  the  resultant 
couple  is  zero  and  the  resultant  is  therefore  zero. 

Motion  under  no  external  forces. —  By  D'Alembert's  principle, 
the  resultant  effective  force  is  zero  if  the  resultant  external  force  is 
zero.  Hence  if  the  external  forces  are  balanced,  the  body  can  have 
no  motion  except  a  uniform  rotation  about  an  axis  containing  the 
center  of  mass.  In  this  case  the  hinge-reaction  is  equal  and  opposite 
to  the  resultant  of  all  other  external  forces,  and  is  zero  if  this  result- 
ant is  zero. 

427.  Unsymmetrical  Body. —  In  the  above  analysis  it  has  been 
assumed  that  the  effective  forces  are  coplanar.  While  it  is  true  that, 
in  the  case  of  rotation  about  a  fixed  axis,  the  effective  forces  are  in 


ROTATION    ABOUT    A    FIXED    AXIS.  359 

parallel  planes,  it  is  not  true  that  they  are  coplanar.  In  certain  cases, 
however,  they  may  be  seen  to  be  equivalent  to  a  coplanar  system. 
This  will  be  true  if  the  body  is  of  uniform  density  and  has  a  plane  of 
symmetry  perpendicular  to  the  axis  of  rotation.  For  if  the  mass  be 
regarded  as  made  up  of  prismatic  elements  whose  axes  are  parallel  to 
the  axis  of  rotation,  the  resultant  effective  force  for  every  element 
will  lie  in  the  plane  of  symmetry.  Even  if  there  is  no  plane  of  sym- 
metry, the  mass  may  be  so  distributed  with  reference  to  a  certain 
plane  that  the  resultant  of  the  effective  forces  lies  in  this  plane.  The 
conditions  which  must  be  satisfied  in  order  that  this  may  be  true 
cannot  here  be  discussed. 

In  the  general  case  of  an  unsymmetrical  body,  the  six  dynamical 
equations  for  forces  in  three  dimensions  are  needed  for  the  complete 
determination  of  the  relations  among  the  external  forces.  The  motion 
may,  however,  in  all  cases  be  determined  from  equation  (i)  of  Art. 
421. 


CHAPTER   XXL 

ANY    PLANE    MOTION    OF    A    RIGID    BODY. 

§  i.   Nature  of  Plane  Motion. 

428.  Coordinates  of  Position. —  Plane  motion  of  a  rigid  body 
has  been  denned  in  Art.  416.  It  was  there  pointed  out  that  in  the 
general  case  of  plane  motion  three  coordinates  serve  to  specify  the 
position  of  the  body  completely. 

In  analyzing  plane  motion  it  is  sufficient  to  consider  a  single  plane 
section  parallel  to  the  motion,  since  the  position  of  every  particle  is 
determined  if  the  positions  of  the  particles  in  this  plane  section  are 
known.  In  the  following  analysis  little  use  is  made  of  coordinates  of 
position,  but  the  geometrical  relations  are  studied  directly.  When 
coordinates  are  employed,  they  will  generally  be  chosen  (as  in  Art. 
416)  as  the  rectangular  coordinates  of  some  particle  A,  and  the  angle 
between  some  fixed  line  and  the  line  joining  two  particles.  The 
point  A  may  be  chosen  arbitrarily,  but  will  usually  be  taken  at  the 
mass-center ;  the  plane  of  the  coordinate  axes  being  so  chosen  as  to 
contain  that  point. 

429.  Displacement. —  The  displacement  of  every  particle  of  a 
rigid  body  (restricted  to  plane  motion)  is  determined  if  the  displace- 
ments of  two  particles  are  known.  The  simplest  displacements  of  a 
body  are  translation  and  rotation. 

The  displacement  is  a  translation  if  all  particles  receive  equal 
and  parallel  displacements.  The  direction  of  the  straight  line  joining 
any  two  points  is  left  unchanged  by  a  translation. 

The  displacement  is  a  rotation  if  every  particle  describes  the  arc 
of  a  circle ;  the  centers  of  the  circles  all  lying  upon  a  straight  line 
called  the  axis  of  rotation.  In  the  following  discussion,  the  atten- 
tion being  directed  to  a  section  of  the  body  by  a  plane  parallel  to  the 
motion,  the  axis  of  rotation  is  represented  by  the  point  in  which  it 
pierces  this  plane.  This  point  is  called  the  center  of  rotation ;  and 
the  terms  center  and  axis  are  often  used  interchangeably. 

The  total  displacement  during  any  interval  is  called  a  translation 
or  a  rotation  if  the  change  from  the  initial  to  the  final  position  can 


PLANE    MOTION    OF    A    RIGID    BODY.  36 1 

be  produced  in  the  manner  specified  in  the  foregoing  definitions, 
whatever  the  series  of  intermediate  positions  through  which  the  par- 
ticles actually  pass. 

430.  General  Plane  Displacement  of  a  Rigid  Body. —  Every 
possible  plane  displacement  of  a  rigid  body  is  equivalent  either  to  a 
rotation  or  to  a  translation. 

The  displacement  is  completely  known  if  the  displacements  of 
two  particles  are  known.  Let  two  par- 
ticles initially  at  A  and  B  (Fig.  172)  be 
displaced  into  the  positions  A'  and  B' . 
Bisect  AA'  at  C  and  BB'  at  B,  and 
draw  from  C  a  line  perpendicular  to 
A  A'  and  from  D  a  line  perpendicular  to 
BB'.  If  these  lines  are  not  parallel,  let 
O  be  their  point  of  intersection.  The 
triangles  OAB  and  OA'B'  are  equal  in 
all  their  parts,  since  by  construction  OA 
=  0A\  OB  =  OB',  AB  =  A'B'.  Therefore  the  angles  A  OB  and 
A' OB'  are  equal,  and  the  angle  AOA'  is 
equal  to  the  angle  BOB'.  It  is  evident  that 
the  triangle  OAB  may  be  brought  into  co- 
incidence with  OA'B'  by  rotation  about  O 
through  an  angle  AOA'. 
~g  Suppose  next  that  A  A'  and  BB'  are  par- 

Fig.  173.  allel.     In  this  case  either  A  A'  and  BB'  are 

equal  (Fig.  173)  and  the  total  displacement  is 
a  translation,  or  else  AB  and  A'B'  are  re- 
lated as  shown  in  Fig.  174,  and  AB  may 
be  brought  to  coincide  with  A'B'  by  a  ro- 
tation about  their  point  of  intersection  0. 
The  proposition  is  thus  true  for  all  cases. 
A  translation  may  be  regarded  as  a 
rotation  about  an  infinitely  distant  axis. 
Thus,  Fig.  173  represents  a  limiting  case  of  \  / 

Fig.  172,  the  lines  CO  and  DO  becoming  v0 

parallel  and  the  point  0  falling  at  infinity.  IG'  I74' 

431.  Instantaneous  Translation  and  Instantaneous  Rotation. — 

The  motion  may  at  any  instant  be  translatory,  even  though  it  does 


A* 

4 

i 

i 

B 

1 

i 

I 

i 

362  THEORETICAL    MECHANICS. 

not  continue  to  be  so  for  any  finite  time.  Again,  the  body  may  at 
any  instant  be  rotating  about  a  certain  axis,  even  though  this  condi- 
tion does  not  continue  for  a  finite  time. 

Instantaneous  translator)'  motion. —  The  instantaneous  motion 
is  translatory  if  the  velocities  of  all  particles  are  equal  in  magnitude 
and  direction  at  the  instant  considered. 

Instantaneous  rotation.  —  The  instantaneous  motion  is  rotational 
if,  at  the  instant  considered,  all  particles  in  a  certain  axis  are  at  rest. 
This  axis  is  called  the  axis  of  instantaneous  rotation,  or  simply  the 
instantaneous  axis.  Every  particle  must  be  moving  at  right  angles 
to  the  perpendicular  drawn  from  it  to  the  instantaneous  axis,  and  the 
velocities  of  different  particles  must  be  proportional  to  their  distances 
from  the  axis. 

432.  Nature  of  Instantaneous  Motion  in  General  Case. — 
Any  possible  plane  motion  of  a  rigid  body  is  at  every  instant  either 
a  translation  or  a  rotation. 

Whatever  the  motion  of  the  body,  let  MM'  and  NN'  (Fig.  175) 
be  the  paths  described  by  two  particles.  Let  A,  B  be  the  positions 
of  the  particles  at  a  certain  instant,  and  A\  B'  their  positions  after  a 
short  interval.     In  general  the  total  displacement  is  equivalent  to  a 

rotation  about  a  center  0,  determined 
by  the  intersection  of  the  perpendicu- 
lar bisectors  of  A  A'  and  BE'  (Art. 
430).  Draw  circular  arcs  AA',  BE', 
/\\v         NN    |\    \  with  common  center  O.     If  the  inter- 

val  be  made  to  approach  zero,  these 
>  circular   arcs    approach    coincidence 

V       with   the   chords  AA\   BE',    which 
themselves  approach  coincidence  with 
2T  the  tangents  to  the  paths  at  A  and 

Fig.  175.  B ;    and   the   point   O  approaches  a 

limiting  position  Q,  determined  by 
the  intersection  of  the  normals  to  the  paths  at  A  and  B.  The  in- 
stantaneous motion  is  therefore  a  rotation  about  Q.  This  point  is 
the  instantaneous  center ;  the  line  through  Q  perpendicular  to  the 
plane  of  motion  is  the  instantaneous  axis. 

In  particular  cases  the  tangents  to  the  paths  at  A  and  B  may  be 
parallel ;  if  the  normals  are  not  coincident,  the  point  Q  is  at  infinity. 
In  such  a  case  the  velocities  of  A  and  B  are  equal  and  parallel  and 


PLANE    MOTION    OF    A    RIGID    BODY.  363 

the  instantaneous  motion  is  translatory.  Translation  is  thus  a  special 
case  of  rotation,  the  instantaneous  axis  being  at  infinity. 

If  the  normals  to  the  paths  at  A  and  B  are  coincident,  their  point 
of  intersection  is  indeterminate.  This  case  may  occur  if  the  motion 
is  translatory  and  A  Bis  perpendicular  to  the  direction  of  the  instan- 
taneous motion  ;  or  if  the  motion  is  a  rotation  about  a  center  lying 
in  AB.  This  case  may  be  avoided  by  replacing  one  of  the  points  B 
by  another  C  such  that  AC  is  not  parallel  to  AB.  (Compare  the 
third  case  of  Art.  430. ) 

As  the  body  moves,  the  position  of  the  instantaneous  center  Q 
(Fig.  175)  in  general  varies  continuously,  both  in  the  body  and  in 
space.  The  curve  described  by  it  is  called  a  centrode ;  the  curve 
traced  in  space  is  the  space  centrode,  and  the  curve  traced  in  the 
moving  body  is  the  body  centrode. 

It  is  evident  from  the  above  discussion  that  the  position  of  the 
instantaneous  center  can  be  determined  if  the  directions  of  motion  of 
two  particles  are  known,  unless  both  are  moving  at  right  angles  to 
the  line  joining  them. 

433.  Angular  Motion  of  a  Body.—  Angular  Displacement. — 
In  any  plane  motion  of  a  rigid  body,  all  lines  fixed  in  the  body  and 
parallel  to  the  plane  of  the  motion  turn  through  equal  angles  in  any 
interval  of  time.  The  angle  turned  through  by  any  such  line  is  called 
the  angular  displacement  of  the  body. 

If  6  is  the  angle  between  a  line  fixed  in  the  body  and  a  line  fixed 
in  space  (as  in  Art.  416),  the  angular  displacement  is  equal  to 
6"  —  0',  6'  being  the  initial  and  6"  the  final  value  of  6. 

Angular  velocity.  —  The  angular  velocity  is  the  angular  displace- 
ment per  unit  time.     Representing  its  value  by  cof  it  is  given  by  the 

formula  <o  =  deidt. 

Angular  acceleration.  —  The  increment  of  the  angular  velocity 
per  unit  time  is  the  angular  acceleration.     Representing  it  by  <f>}  its 

value  is  <f>  =  dco/dt  =  d>0/dt\ 

These  formulas  for  the  angular  motion  have  the  same  form  in  the 
general  case  of  plane  motion  as  in  the  particular  case  of  rotation 
about  a  fixed  axis  (Art.  415). 

434.  Motions  of  Individual  Particles.— If  at  any  instant  the 
angular  velocity  and  the  instantaneous  axis  are  known,  the  velocity 


364 


THEORETICAL    MECHANICS. 


of  every  particle  is  known.  If  r  is  the  perpendicular  distance  of  the 
particle  from  the  instantaneous  axis  and  &>  the  angular  velocity,  the 
particle  has  a  velocity  rco  at  right  angles  to  r. 

If  vx  is  the  velocity  of  a  particle  A  and  v2  that  of  a  particle  B, 
the  resolved  parts  of  vx  and  v2  in  the  direction  AB  are  equal  at 
every  instant.  For  otherwise  the  distance  AB  would  not  remain 
constant. 

Examples. 

1.  The  ends  of  a  straight  bar  slide  along  intersecting  lines  at  right 
angles  to  each  other.  Determine  the  instantaneous  center  at  any 
instant. 

2.  If,  in  Ex.  1,  the  length  of  the  bar  is  a  and  its  angular  velocity 
a),  determine  the  velocity  of  each  end  and  of  the  middle  point  at  the 
instant  when  the  length  of  the  bar  coincides  with  one  of  the  guiding 
lines. 

3.  Let  the  length  of  the  bar  be  1 2  ins.  and  its  angular  velocity 
constantly  5000  per  sec.  Determine  the  velocity  of  each  end  and  of 
the  middle  point  when  the  bar  makes  an  angle  of  6o°  with  one  of  the 
guiding  lines. 

Am.  Velocity  of  middle  point  =  4.36  ft. -per-sec. ,  directed  at 
angle  300  with  bar. 

4.  One  end  of  a  bar  describes  a  circle,  the  other  moves  along  a 
straight  line  containing  a  diameter  of  the  circle.  Determine  the  in- 
stantaneous center  in  each  of  the  two  positions  in  which  the  bar 
coincides  with  the  directing  straight  line. 

5.  In  Ex.  4,  let  a  =  length  of  bar,  r  =  radius  of  circle,  v  =  ve- 
locity of  the  point  describing  the  circle.  Determine  the  angular 
velocity  of  the  bar  when  it  makes  an  angle  0  with  the  guiding  straight 
line.  Ans.   ft)  =  (V2  —  a%  sin2  0)1Av/ar  cos  6. 

6.  In  Ex.  5,  let  A  be  the  point  describing  the  straight  line,  B 
the  point  describing  the  circle,  C  the  center  of  the  circle.  Determine 
the  velocity  of  the  point  A  when  A  CB  is  a  right  angle.  Determine 
the  velocity  of  the  middle  point  of  the  bar  at  the  same  instant. 

7.  Where  is  the  instantaneous  center,  and  what  is  the  angular 
velocity,  when  the  bar  is  in  the  position  described  in  Ex.  6  ? 

§  2.    Composition  and  Resolution  of  Plane  Motions. 

435.  Component  and  Resultant  Motions  of  a  System  of  Par- 
ticles.—  Just  as  it  is  often  useful,  in  the  analysis  of  the  motion  of  a 
single  particle,  to  regard  its  actual  velocity  as  made  up  of  compo- 
nents, so  in  studying  the  motion  of  a  system  of  particles  it  may  be 


PLANE    MOTION    OF    A    RIGID    BODY.  365 

useful  to  regard  any  actual  instantaneous  motion  as  the  resultant  of 
two  or  more  component  motions.  The  meaning  of  component  and 
resultant  motions  of  a  system  may  be  defined  as  follows : 

Any  definite  instantaneous  motion  of  a  system  of  particles  implies 
definite  velocities  of  all  the  particles.  Conceive  two  such  motions 
M\  M"  ;  let  vXi  vj,  v3',  ...  be  the  velocities  of  the  individ- 
ual particles  in  the  motion  M' ,  and  tr",  p*t  v./,  .  .  .  their 
velocities  in  the  motion  M" .  Let  vx  be  the  vector  sum  of  vx  and 
v{%  v2  the  vector  sum  of  v2  and  v2",  v.6  the  vector  sum  of  v^  and  v3", 
.  .  .  ;  and  designate  by  Ma.  motion  of  the  system  in  which  the 
individual  particles  have  velocities  vu  v2,  v$t  ....  Then  the 
motion  M  may  be  regarded  as  the  resultant  of  the  motions  M'  and  M" . 

436.  Component  and  Resultant  Motions  of  a  Rigid  Body. — 

The  velocities  of  the  individual  particles  of  a  rigid  body  are  not  in- 
dependent of  one  another.  In  plane  motion  the  velocity  of  every 
particle  is  determined  if  the  velocities  of  two  particles  are  assigned, 
and  even  these  two  velocities  cannot  be  assigned  arbitrarily,  but  must 
satisfy  the  condition  that  their  resolved  parts  along  the  line  joining 
them  are  equal  (Art.  434).  Using  the  same  notation  as  above,  let 
M'  and  M"  be  two  possible  instantaneous  motions  of  a  rigid  body ; 
then  it  may  be  shown  that  the  resultant  motion  M  is  a  possible 
motion  of  the  rigid  body. 

In  order  to  prove  this,  it  is  sufficient  to  show  that  the  particles 
may  possess  velocities  Vxtv%9V%i  .  .  .  without  changing  their 
distances  apart ;  in  other  words,  that  the  resolved  parts  of  any  two 
velocities  vt,  v2  along  the  line  joining  the  two  particles  are  equal. 
That  this  condition  is  satisfied  by  the  velocities  vx ,  v2  follows  at  once 
from  the  fact  that  it  is  satisfied  by  v/,  v2\  and  also  by  v"%  v2  ;  and 
the  fact  that  vx  is  the  vector  sum  of  vx\  vx\  and  v2  the  vector  sum  of 
v2,  v2.      In  general,  therefore, 

The  resultant  of  any  two  instantaneous  motions  of  a  rigid  body 
is  a  motion  consistent  zvit/i  rigidity. 

Any  actual  motion  of  a  rigid  body  may  therefore  be  regarded  as 
the  resultant  of  two  or  more  simultaneous  motions.  It  remains  to 
consider  the  principles  governing  the  composition  and  resolution  of 
rigid-body  motions. 

437.  Resultant  of  Two  Instantaneous  Motions. —  It  has  been 
shown  that  any  possible  plane  motion  of  a  rigid  body  is  at  every 


366  THEORETICAL    MECHANICS. 

instant  a  rotation  about  a  definite  axis  (reducing  in  a  particular  case 
to  a  translation).  The  resultant  of  two  given  motions  may  be  deter- 
mined as  follows : 

Let  the  component  motions  be  rotations  about  instantaneous 
centers  C  and  C" ,  with  angular  velocities  co'  and  co";  and  let  the 
resultant  motion  be  a  rotation  about  an  instantaneous  center  C  with 
angular  velocity  co.  It  is  required  to  determine  the  position  of  C 
and  the  value  of  co.  Consider  separately  the  cases  in  which  co'  and 
to"  have  (a)  the  same  and  (^)  opposite  angular  directions. 

(a)  The  resultant  velocity  of  a  particle  at  the  instantaneous  center 
C  is  zero ;  its  two  components  due  to  the  two  given  rotations  are 

therefore   equal   and    opposite.     But 
.«-.       /*~^v  /*~"\  two  comPonent  velocities  are  not 

f  C    )    f   C    ) (    fr*  \     parallel  for  any  particle  not  in  the  line 

\J-^Zn\^^-"-^--\^_y      C  C"  \  hence  C  must  lie  in  that  line. 

Fig.  176.  Again,  if  co  and  co"  have  the  same 

angular   direction    (Fig.    176),    it  is 

only  for  a  particle  between   C  and   C"  that  the  two  component 

velocities   have   opposite  directions ;    C  therefore   lies   between  C 

and  C" . 

Let  .a',  z"  denote  the  distances  C C,  C" C  respectively.  A  par- 
ticle at  C  has  a  velocity  z'co'  due  to  the  rotation  about  C,  and  an 
opposite  velocity  z"co"  due  to  the  rotation  about  C" \  and  since  its 
resultant  velocity  is  zero, 

z'co'  =  z"co"     or     z'/z"  =  co' '/co' '. 

The  value  of  co,  the  angular  velocity  of  the  resultant  motion,  may 
be  found  by  determining  the  velocity  of  any  particle  not  at  C.  The 
two  components  of  velocity  of  a  particle  at  C  are  o  and  {z'  -\-  z")co"; 
its  resultant  velocity  is  therefore  {z'  -\-  z")co",  and  the  angular  ve- 
locity of  the  line  CC  is 

co  =co"(z'  +  z")/z'; 

or,  substituting  the  above  value  of  z'/z", 

co  =  co'  -\-  co". 

(b)  In  case  the  angular  directions  of  co'  and  co"  are  opposite,  let 
C  and  C"  (Fig.  177)  represent  the  two  instantaneous  centers,  and 
let  it  be  assumed  that  co  is  greater  than  co". 


PLANE    MOTION    OF    A    RIGID    BODY.  367 

As  in  the  former  case,  it  is  seen  that  C  must  lie  on  the  line  C  C'\ 
its  position  being  such  that  the  components  of  velocity  of  a  particle 
at  C  due  to  the  two  rotations  shall  be  equal  and  opposite.  The  com- 
ponents will  not  have  opposite  directions  for  particles  between  C 
and  C"\  and  since  ft)'  is  greater  than  ft)",  the  two  components  of 
velocity  can  be  equal  only  if  CC  is 
less  than  CC".  The  point  (7  there- 
fore lies  without  the  length  C  C  in  *%1  «J*  «* 

the  direction  of  C.    Its  position  must       f  C_\ f    ff    1  [     y\ 

be  such  that  Vj^/  \J~^4' v!"*^/ 

z'co'  =  z"(o",     or     *'/#*  =  o)"/c»'.  pIG   x 

To  determine  the  angular  velocity 
of  the  resultant  motion,  notice  that  the  resultant  velocity  of  C  is  the 
sum  of  components  o  and  (z"  —  z')(o"  ;  the  angular  velocity  of  the 
line  CC  is  therefore 

ft)  =  (o"(z"  —  -O/^'  ===  m>  —  w  • 

Case  of  two  equal  and  opposite  angular  velocities. — If,  in  the  case 
in  which  ft)'  and  ft)"  have  opposite  directions,  their  magnitudes  are 
made  to  approach  equality,  the  point  C  recedes  from  C,  and  in  the 
limit,  when  ft)'  =  ft)",  C  passes  to  infinity.  In  this  case  ft)  =  o ;  that 
is,  the  system  has  no  angular  velocity,  and  the  motion  is  a  transla- 
tion. The  translational  velocity  is  equal  to  the  velocity  of  any  particle 
as  C,  that  is,  it  is  perpendicular  to  C '  C"  and  has  the  value 

O"  —  *')*"  =  (CC)co". 

438.  Analogy  to  Parallel  Forces. —  The  results  of  the  above 
discussion  show  a  close  analogy  between  the  composition  of  rotations 
and  the  composition  of  parallel  forces.  Thus,  if  C  and  C"  ( Fig. 
176  or  Fig.  177)  are  the  points  of  application  of  parallel  forces  of 
magnitudes  ft)'  and  ft)",  their  resultant  is  a  force  of  magnitude  ft)  ap- 
plied at  C;  and  ft)  is  equal  to  ft)'  -f-  ft)"  or  to  ft)'  —  ft)",  according  as 
the  two  forces  have  the  same  direction  or  opposite  directions. 

The  case  of  two  equal  and  opposite  rotations  is  analogous  to  that 
of  a  couple  ;  thus  a  translation  may  be  regarded  mathematically  as  a 
rotation  couple. 

The  analogy  may  without  difficulty  be  shown  to  hold  for  any 
number  of  component  motions.  The  process  of  finding  the  angular 
velocity  and  instantaneous  axis  of  the  resultant  is  identical  with  the 


368  THEORETICAL    MECHANICS. 

process  of  finding  the  magnitude,  direction  and  line  of  action  of  a 
system  of  parallel  forces.  The  lines  of  action  of  the  component  forces 
coincide  with  the  instantaneous  axes  of  the  component  rotations  ;  the 
magnitude  of  any  one  of  the  forces  is  equal  to  the  angular  velocity  of 
the  corresponding  motion  ;  and  opposite  rotation-directions  are  rep- 
resented by  opposite  force-directions. 

439.  Resolution  of  a  Rotation  Into  Components. —  By  reversing 
the  above  process,  the  actual  motion  of  a  body  at  any  instant  may 
be  resolved  into  components. 

Examples. 

1 .  What  single  motion  of  a  rigid  body  is  equivalent  to  rotational 
velocities  of  10  rad. -per-sec.  and  4  rev. -per-sec.  in  the  same  direc- 
tion, about  instantaneous  centers  5  ft.  apart? 

2.  Two  rotations  of  5  rad. -per-sec.  and  8  rad. -per-sec.  in  oppo- 
site directions,  about  instantaneous  centers  4  ft.  apart,  are  equivalent 
to  what  single  rotation  ? 

3.  What  is  the  resultant  of  two  rotations  of  10  rev.-per-min.  in 
opposite  directions,  about  centers  2  ft.  apart? 

4.  A  rotation  of  4  rev.  -per-sec.  about  a  center  C  is  equivalent  to 
what  two  rotations  about  centers  distant  1  ft.  and  2  ft.  from  C  on  op- 
posite sides  ? 

5.  A  rotation  of  5  rev. -per-sec.  about  a  center  C  is  equivalent  to 
what  two  rotations  about  centers  distant  3  ft.  and  2  ft.  from  C  in  the 
same  direction? 

6.  A  translational  velocity  of  20  ft. -per-sec.  is  equivalent  to  what 
angular  velocities  about  two  points  8  ft.  apart?  What  must  be  the 
direction  of  the  line  joining  the  proposed  centers  in  order  that  the 
resolution  may  be  possible? 

7.  Show  how  to  resolve  a  given  rotation  into  three  components 
whose  centers  are  given  and  are  not  collinear. 

440.  Translation  and  Rotation. —  The  above  processes  of  com- 
bining or  resolving  rotations  include  the  case  in  which  one  of  the 
components  is  a  translation.  An  independent  treatment  of  this  case 
is,  however,  desirable. 

Composition. —  Let  the  resultant  motion  of  a  rigid  body  at  a 
certain  instant  be  made  up  of  two  components,  one  of  which  is  a 
rotation  about  the  instantaneous  center  C  (Fig.  178)  with  angular 
velocity  ft),  and  the  other  a  translation  of  velocity  v  in  the  direction 
CM.     Since  this  translational  velocity  is  the  resultant  velocity  of  C, 


PLANE    MOTION   OF    A    RIGID    BODY.  369 

the  instantaneous  center  C  of  the  resultant  motion  must  lie  on  a  line 
through  C  perpendicular  to  CM.     If  z  de- 
notes the  distance  CC,  we  must  have 

z(a  =  v,     or     z  =  v/tOy 
since  the  resultant  velocity  of  a  particle  at 
C  must  be  zero.* 

Resolution. —  By  reversing  the  above 
process,  any  given  motion  may  be  resolved 
into  a  translation  and  a  rotation. 

As  a  special  case,  any  motion  may  be  ~Fig  173 

resolved  into  a  translation  and  a  rotation 

about  an  instantaneous  center  chosen  at  pleasure.  To  accomplish 
this,  it  is  only  necessary  to  select  as  the  translational  component  of 
velocity  for  every  particle  the  velocity  of  the  point  which  is  to  be 
made  the  instantaneous  center.  This  will  be  illustrated  in  treating 
the  dynamics  of  plane  motion  of  a  rigid  body. 

Examples. 

1.  A  rotation  of  ioo  rev. -per-min.  is  to  be  resolved  into  two 
components,  one  of  which  shall  be  a  translation  equal  to  the  actual 
velocity  of  a  certain  particle  4  ft.  from  the  center  of  rotation.  De- 
termine the  two  components  completely. 

2.  A  rotation  of  100  rev. -per-min.  about  a  center  C  and  a  trans- 
lation of  86  ft. -per-min.  in  a  certain  direction  are  equivalent  to  what 
resultant  motion  ? 

3.  Resolve  a  rotation  of  50  rad. -per-sec.  into  two  components, 
one  of  which  shall  be  a  translation  and  the  other  a  rotation  about  a 
center  6  ft.  from  the  center  of  the  given  rotation. 

Ans.  The  translational  velocity  is  300  ft. -per-sec.  at  right  angles 
to  the  line  joining  the  two  centers  of  rotation. 

441.  Composition  and  Resolution  of  Accelerations. —  If,  in  two 
motions  M ',  M"  of  a  rigid  body,  a  certain  particle  has  accelerations 
p\  p"  respectively,  its  acceleration  in  the  resultant  motion  is  the  vec- 
tor sum  of p'  and/".  This  follows  from  Art.  436,  since  acceleration 
is  change  of  velocity  per  unit  time.  A  particular  case  of  the  resolu- 
tion of  accelerations  is  important  in  the  application  of  D'Alembert's 
principle  to  the  general  case  of  plane  motion  (Art.  443). 

*  It  will  be  noticed  that  the  process  of  combining  a  rotation  and  a  trans- 
lation is  exactly  similar  to  the  process  of  combining  a  force  and  a  couple, 
explained  in  Art.  94. 

24 


370  THEORETICAL    MECHANICS. 

If  A  and  B  are  any  two  particles  of  a  rigid  body  having  any 
plane  motion,  and  if  the  acceleration  of  B  is  resolved  into  two  com- 
ponents, one  of  which  is  equal  to  the  acceleration  of  A,  the  other 
component  is  equal  to  the  acceleration  which  B  would  have  if  the 
body  rotated  about  a  fixed  axis  through  A  with  its  actual  angular 
motion. 

Let  v  be  the  velocity  of  A'  and  p  its  acceleration  ;  then  the  truth 
of  the  proposition  becomes  evident  by  considering  that  if,  with  the 
actual  motion,  there  be  compounded  a  translation  with  velocity  — v 
and  acceleration  — p,  the  particle  A  is  reduced  to  rest  while  the  an- 
gular motion  is  unchanged. 

§  3.  Dynamics  of  Plane  Motion. 

442.  Application  of  D'Alembert's  Principle. —  The  way  in  which 
the  condition  of  motion  of  a  body  is  changing  at  any  instant  depends 
upon  the  shape  and  size  of  the  body,  the  distribution  of  its  mass, 
and  the  forces  acting  upon  every  particle.  The  influence  of  these 
several  elements  may  be  expressed  by  algebraic  equations,  derivable 
from  the  general  principle  known  as  "D'Alembert's  principle."* 
The  application  of  this  principle  to  the  general  plane  motion  of  a 
rigid  body  must  now  be  considered. 

443.  Effective  Forces. —  In  order  to  determine  the  effective 
forces,  let  the  actual  motion  be  resolved  into  two  component  motions 

in  accordance  with  the  principle  stated  in 
Art.  441,  and  let  the  effective  forces  for 
the  component  motions  be  computed 
separately. 

One  of  the  component  motions  is  a 
■171V0O*  rotation  about  a  fixed  axis  with  the  ac- 

tual angular  velocity  and  angular  acceler- 
ation of  the  body ;  the  other  is  a  transla- 
tion with  velocity  and  acceleration  equal 
P  to  those  of  a  particle  lying  in  the  assumed 

Fig.  179.  fixed  axis.    The  position  of  the  fixed  axis 

may  be  chosen  at  pleasure  ;  let  it  be  at  A 
(Fig.  179),  and  let  v  be  the  velocity  and^>  the  acceleration  of  a  par- 

*See  Arts.  387,  390. 


->. 


PLANE    MOTION    OF    A    RIGID    BODY.  37 1 

tide  of  the  body  which  instantaneously  coincides  with  A.  Let  0  be 
the  angular  coordinate  of  the  body  (Art.  433),  co  =  dOjdt  its  angu- 
lar velocity,  and  cj>  =  dco/dt  =  d'26/dt2  its  angular  acceleration.  The 
two  components  into  which  the  actual  motion  of  the  body  is  resolved 
are  then  (a)  a  translation  with  velocity  v  and  acceleration  p ;  (b)  a 
rotation  about  a  fixed  axis  through  A  with  angular  velocity  co  and 
angular  acceleration  <f>.  Consider  the  effective  force  on  any  particle 
due  to  each  of  these  component  motions.  Let  B  (Fig.  179)  repre- 
sent the  position  of  a  particle  of  mass  m}  and  let  AB  =  r. 

{a)  Corresponding  to  the  translation,  the  effective  force  is  equal 
in  magnitude  and  direction  to  mp. 

{b)  Corresponding  to  the  rotation,  the  effective  force  is  made  up 
of  a  component  mrco2  in  direction  BA  and  a  component  mrc\>  at  right 
angles  to  BA. 

Since  the  translational  effective  forces  on  different  particles  are 
proportional  to  their  masses  and  have  a  common  direction,  their  re- 
sultant is  a  force  Mp  applied  at  the  mass-center,  M  being  the  whole 
mass  of  the  body. 

The  resultant  of  the  rotational  effective  forces  may  be  determined 
as  in  Art.  426.  The  moment  of  this  resultant  about  the  axis  of  rota- 
tion is  I<f),  I  being  the  moment  of  inertia  about  that  axis  (Art.  420). 

The  rotational  component  of  the  motion  takes  place  about  an 
axis  chosen  at  pleasure.  In  order  to  simplify  the  system  of  effective 
forces,  let  this  axis  contain  the  mass-center.  Then,  as  shown  in  Art. 
426,  the  rotational  effective  forces  are  equivalent  to  a  couple.  The 
moment  of  this  couple  about  any  axis  whatever  is  equal  to  I(f>  if  /  is 
the  moment  of  inertia  with  respect  to  the  central  axis. 

The  results  of  the  discussion  may  be  summarized  as  follows: 

In  any  plane  motion  of  a  rigid  body,  the  system  of  effective 
forces  is  equivalent  to  a  force  Mp  applied  at  the  mass-center  and  a 
couple  of  moment  7(f) ;  where  M  is  the  total  mass  of  the  body,  /  its 
moment  of  inertia  with  respect  to  a  central  axis  perpendicular  to  the 
plane  of  the  motion,/  the  acceleration  of  the  mass-center,  and  cp  the 
angular  acceleration  of  the  body. 

444,  Determination  of  Angular  Motion. —  By  D'Alembert's 
principle,  the  sum  of  the  moments  of  the  external  forces  about  any 
axis  is  equal  to  the  sum  of  the  moments  of  the  effective  forces  about 
that  axis.  Let  the  axis  of  moments  contain  the  center  of  mass  ;  then 
the  moment  of  the  resultant  of  the  translational  effective  forces  is 


372  THEORETICAL    MECHANICS. 

zero,  and  the  sum  of  the  moments  of  the  effective  forces  is  equal  to 
1(f),  if  /  is  the  moment  of  inertia  with  respect  to  the  central  axis. 
Let  L  denote  the  sum  of  the  moments  of  the  external  forces  about 
this  axis ;  then 

L  =  /<f>. 

This  is  identical  with  the  equation  of  angular  motion  about  a  fixed 
axis  containing  the  mass -center, 

445.  Independence  of  Translation  and  Rotation. —  Any  actual 
motion  of  a  body  being  regarded  as  made  up  of  a  rotation  about  an 
axis  containing  the  mass-center,  together  with  a  translation  equal  to 
that  of  the  mass-center,  we  have  now  proved  that  these  two  motions 
take  place  independently.  The  meaning  of  this  statement  may  be 
expressed  definitely  as  follows  : 

( 1 )  The  acceleration  of  the  mass-center  is  the  same  as  if  the  whole 
mass  were  concentrated  at  that  point  and  acted  upon  by  forces  equal 
in  magnitude  and  direction  to  the  actual  external  forces. 

(2)  The  angular  acceleration  is  the  same  as  if  the  mass-center 
were  fixed  and  the  actual  external  forces  applied. 

446.  Complete  Determination  of  the  Motion. —  In  the  general 
case  of  plane  motion,  three  independent  dynamical  equations  may  be 
written.  Two  of  these  may  be  the  equations  of  motion  of  the  mass- 
center,  the  third  the  equation  of  angular  motion.  In  any  determinate 
problem  these  three  equations,  together  with  the  initial  conditions 
for  determining  constants  of  integration,  and  the  geometrical  relations 
(if  the  motion  is  constrained)  serve  to  determine  the  motion  and  the 
unknown  forces. 

Let  the  coordinates  of  the  body  be  taken  as  in  Art.  428.  Choosing 
a  pair  of  rectangular  axes  lying  in  a  plane  parallel  to  the  motion  and 
containing  the  mass-center,  let 

x,  y  =  coordinates  of  mass-center  ; 
6  =  angular  coordinate  of  body ; 
M  =  total  mass  of  body  ; 
/==  Mk2  =  moment  of  inertia  with  respect  to  central  axis; 
P  =  vector  sum  of  external  forces  ; 
X,  Y  =  axial  components  of  P ; 

L  =  sum  of  moments  of  external  forces  about  central  axis. 


PLANE    MOTION   OF    A    RIGID    BODY. 


373 


The  three  equations  of  motion  take  the  form 

M(d2x/dt2)  =  X ;     ....  (i) 

M{<Pyldt*)  =  Y;    .         .        .        .  (2) 

Kd*6/dt*)  =  L (3) 

447.  Applications. —  The  methods  of  applying  the  above  prin- 
ciples will  be  illustrated  by  the  solution  of  the  following  problems  : 

I.  Determine  the  motion  of  a  homogeneous  circular  cylinder 
which  rolls,  without  sliding,  on  an  inclined  plane,  under  the  action  of 
gravity. 

Solution. —  Let  /3  =  inclination  of  plane  to  horizon  ;  a  =  radius 
of  cylinder ;  M  =  its  mass ;  k  =  radius  of  gyration  about  its  geo- 
metrical axis  ;  {kl  =  \dl  by  Art.  398, 
Ex.  1  ;)  x,  y,  6  =  coordinates  of  po- 
sition, the  origin  and  axes  being  taken 
as  shown  in  Fig.  180. 

The  external  forces  acting  upon 
the  cylinder  are  its  weight  and  the  re- 
action of  the  plane.  The  weight  is 
equivalent  to  a  force  Mg,  acting  ver- 
tically downward  at  the  mass-center. 
The  reaction  of  the  plane  is  unknown  in  magnitude  and  direction, 
but  may  be  replaced  by  its  normal  and  tangential  components  N  and 
T,  as  shown  in  the  figure.  Noticing  that  the  /-component  of  accel- 
eration of  the  mass-center  is  zero,  since  y  is  constant,  the  three  equa- 
tions may  be  written  as  follows  : 

M(d*x/dt2)  =  Mg  sin  £  —  T;         .         .     (1) 

o  =  Mg  cos  fi  —  N ;       .         .     (2) 

iMdXd20/dn  =  Ta (3) 

Since  the  cylinder  rolls  without  sliding,  there  is  also  the  geometrical 
equation 

x  =  a6  -f-  constant, 


Fig.  180. 


from  which 


d^xldt1  =  a(d'20/df2). 


(4) 


These  equations  may  be  solved  as  follows: 

Substituting   in  (1)  the  value  of  T  from  (3)  and  the  value  of 
dlx\dtl  from  (4),  and  reducing, 

d'l6ldt2  =  (2g  sin  /3)/3<?. 


374  THEORETICAL    MECHANICS. 

Integrating  and  assuming  that  6  and  dO/dt  are  both  zero  when  t  =  ot 

d  =  \_(gsml3)!?>a]t\  .         .         .      (5) 

This  determines,  the  motion.  If  the  origin  is  so  taken  that  x  =  o 
when  6  =  o, 

x=  ad  =  igsin  $  ■  t\  .         .         .     (6) 

To  complete  the  solution  the  values  of  N  and  T  must  be  found. 
From  (i), 

T  =  M(g  sin  j3  —  x)  =  iMg  sin  /3.  .  .     (7) 

From  (2), 

N  =  Mg  cos  /3.         .       : .         .         .     (8) 

II.   A  homogeneous  circular  cylinder  is  placed  with  its  axis  ver- 
tical on  a  smooth  horizontal  plane,  and  is  set  in  motion  by  a  con- 
stant tension  applied  to  the  free  end 
■*  1  of  a   flexible    cord    wound    upon    its 

eK  surface.     Determine  the  motion. 

0\\  Solution.  —  Let  M  =  mass  of  cyl- 

7  X       inder  ;  a  =  its  radius  ;  ^  =  its  radius 

y  of  gyration  with  respect   to  its  geo- 

~^T         metrical  axis  ;  (/P  =  \a2 ;)  /  =  Mk2  == 
Fig.  181.  \Ma2 ;     T=  tension   in   string;  x, 

j/,  6  =  coordinates  of  position  taken 
in  the  usual  way,  the  axes  being  as  shown  in  Fig.  181. 

Since  the  only  external  force  acting  on  the  cylinder  in  the  plane 
of  the  motion  is  T,  the  acceleration  of  the  mass-center  has  the  direc- 
tion of  T.     The  dynamical  equations  are 

M(d2x/dt2)  =  T\        .         .         .         .     (1) 
\Ma\d26ldt2)  =  Ta (2) 

If  T  is  a  known  constant,  the  motion  can  be  determined  by  the  inte- 
gration of  these  two  equations.  If  T  is  unknown,  the  motion  cannot 
be  completely  determined  ;  but  in  any  case  the  elimination  of  T 
shows  that  the  angular  acceleration  and  the  linear  acceleration  bear  a 
constant  ratio  to  each  other  ;  that  is, 

d2x/dt2  =  \a{d20jdt2). 

Examples. 

1.  A  homogeneous  cylinder  2  ft.  in  diameter  rolls  without  sliding 
down  a  plane  inclined  300  to  the  horizon,      (a)  Determine  the  dis- 


PLANE    MOTION   OF   A   RIGID    BODY.  375 

tance  moved  by  the  mass-center  in  2  sec. ,  starting  from  rest,  (b)  If 
the  mass  of  the  cylinder  is  40  lbs.,  determine  the  normal  and  tangen- 
tial components  of  the  reaction  of  the  plane  on  the  cylinder. 

Ans.  (a)  21.5  ft,  nearly.  (&)  T  =  one-sixth  the  weight  of  the 
body. 

2.  Solve  the  problem  of  the  motion  of  a  cylinder  rolling  on  an 
inclined  plane,  assuming  that  the  body  has  initially  a  motion  up  the 
plane.  Apply  the  results  to  the  case  described  in  Ex.  1,  assum- 
ing an  initial  velocity  of  the  mass-center  of  20  ft-per-sec.  up  the 
plane. 

Ans.  The  body  will  come  to  rest  at  the  end  of  bolg  sec,  and  will 
then  descend  as  in  Ex.  1. 

3.  Determine  the  motion  of  a  homogeneous  sphere  which  rolls 
without  sliding  down  a  rough  inclined  plane,  and  apply  the  results  to 
a  sphere  of  10  lbs.  mass  and  2  ins.  diameter  rolling  from  rest  down  a 
plane  inclined  50  to  the  horizon. 

Ans.  The  acceleration  of  the  mass-center  is  5/7  that  of  a  body 
sliding  on  the  plane  without  friction. 

4.  A  homogeneous  sphere  rolls,  without  sliding,  on  the  inner  sur- 
face of  a  hemisphere.  If  it  moves  from  rest  under  the  action  of  no 
force  except  gravity  and  the  reaction  of  the  surface,  determine  the 
motion. 

Let  A  be  the  center  of  the  hemisphere  (Fig.  182),  A  that  of  the 
sphere,  B  the  lowest  position  of  A,  AE  that  radius  of  the  sphere 
which  is  vertical  when  A  A'  is  vertical. 
Let  a  =  radius  of  sphere,  a '  =  radius 
of  hemisphere,  <f>  =  angle  AA'B,  6  = 
angle  between  AE  and  vertical. 

In  writing  the  equations  of  motion 
of  the  mass-center,  let  the  acceleration 
be  resolved  along  the  tangent  and  nor- 
mal to  the  circle  described  by  A.  The 
components  are  {a'  —  a)(d2<f>/dt2)  and 
(a'  —  a)(d<\>\d?jl.    Let  the  pressure  on  Fig.  182. 

the  sphere  at  the  point  of  contact  D  be 

resolved  into  a  normal  component  N  and  a  tangential  component  T, 
both  unknown.     The  only  other  external  force  is  the  weight  Mg. 

The  equations  of  motion  are 


T  —  Mg  sin  <j>  =  M{a '  —  a)(d2(f>ldt2);    . 

•     (0 

N—  Mg  cos  $  =  M(a'  —  d){d$ldf? ;      . 

•     (2) 

—  Ta  =  (2Ma2/5Xd26/dt2). 

•     (3) 

The  angles  0  and  <f>  are  related  in  a  simple  manner. 

Since  the 

arcs  ED  and  CD  are  equal,  and  angle  DAE  =  0  -\-  (/>, 

a'<t>  =  a{6  +  </>).     . 

•  (4) 

376  THEORETICAL    MECHANICS. 

From  equation  (4), 

a{d20ldt2')  =  («'  —  d)(d24>ldt*). 
Substituting  this  value  in  (3),  eliminating  T  between  (3)  and  (1),  and 

reducing,  dj±  = k_sin£        .  .  .      (5) 

dt2  7<y  —  a)  ^ 

This  equation  is  identical  in  form  with  that  for  the  motion  of  a  pen- 
dulum. The  complete  solution  involves  an  elliptic  integral,  but  one 
integration  can  be  performed  as  in  Art.  425.  If  the  sphere  is  at  rest 
when  cf>  =  <£0, 

(?)  '=  -^-, (cos  *- cos  *•>•    •    •  (6) 

\dtj       y{af  —  a) 
From  (1)  and  (5), 

T=(2Mgs\n$)h.  .         .         .     (7) 

From  (2)  and  (6), 

N  =  Mg(\i  cos  </>  —  10  cos  4>o)l7-  -         •     (8) 

5.  In  Ex.  4,  how  great  must  the  coefficient  of  friction  be  in  order 
that  the  assumed  rolling  without  sliding  may  be  possible  ? 

The  coefficient  of  friction  must  be  not  less  than  the  greatest  value 
of  T/N  found  in  the  above  solution.     Let 

/x  =  TIN  =  (2  sin  <£)/(  1 7  cos  c/>  —  10  cos  <£0). 
The  value  of  /x  increases  with  (f>,  and  has  its  greatest  value  when 
(f>  =  <f)0 ;  in  this  position 

p  =  (2  tan  0o)/7, 
which  is  the  least  value  of  the  coefficient  of  friction  compatible  with 
the  assumed  motion. 

6.  In  Ex.  4,  if  the  sphere  oscillates  through  a  small  angle  about 
its  lowest  position,  determine  the  length  of  an  equivalent  simple  pen- 
dulum. Ans.  7 (a'  —  a)j$. 

7.  A  sphere  rolls,  without  sliding,  on  the  outer  surface  of  a 
sphere.  If  it  starts  from  rest  at  the  highest  point,  where  will  it  leave 
the  surface  ?     (Put  the  normal  reaction  equal  to  zero. ) 

Ans.  Let  (f>  =  angle  between  vertical  and  line  joining  centers ; 
then  the  spheres  separate  when  cos  <j>  =  10/17  ;  (j>  —  530  58'. 

8.  In  Ex.  7,  if  the  spheres  are  smooth,  show  that  they  will  sep- 
arate when  cos  </>  =  2/3. 

9.  A  circular  cylinder  whose  mass-center  is  not  in  its  geometrical 
axis  rolls,  without  sliding,  upon  a  horizontal  plane.  Let  a  =  radius 
of  circular  section,  c  =  distance  of  mass-center  from  geometrical 
axis,  k  =  radius  of  gyration  with  respect  to  axis  through  mass-center, 
0  =  angular  displacement  from  position  of  stable  equilibrium  at  time/. 

Show  that    ^  ^  a(d6/dty](a2  +  &  +  k2  —  2ac  cos  0) 
remains  constant  during  the  motion. 


PLANE    MOTION    OF    A    RIGID    BODY.  377 

10.  A  circular  cylinder  whose  mass-center  is  not  in  its  geomet- 
rical axis  is  placed  on  a  smooth  horizontal  plane.  With  notation  as 
in  Ex.  9,  show  that 

(£*+  C*  sin2  6)(d0idt)2  —  2gc  cos  d 
remains  constant  during  the  motion. 

1 1.  How  great  must  the  coefficient  of  friction  be  in  order  that  a 
homogeneous  cylinder  placed  on  an  inclined  plane  with  axis  hori- 
zontal shall  roll  without  sliding  ? 

448.  General  Method  of  Writing  Equations  of  Motion. —  The 

three  equations  of  motion  are  not  necessarily  written  as  in  Art.  446, 
although  this  method  furnishes  the  simplest  solution  of  many  prob- 
lems. 

The  equations  derivable  from  D'Alembert's  principle  are  of  two 
kinds,  —  equations  of  resolution  and  equations  of  moments.  All  such 
equations  are  included  in  the  two  following  statements : 

(a)  The  sum  of  the  resolved  parts  of  the  effective  forces  is  equal 
to  the  sum  of  the  resolved  parts  of  the  external  forces,  for  any  direc- 
tion of  resolution. 

(J?)  The  sum  of  the  moments  of  the  effective  forces  is  equal  to 
the  sum  of  the  moments  of  the  external  forces,  for  any  axis. 

By  reasoning  analagous  to  that  used  in  Art.  104  in  discussing 
the  conditions  of  equilibrium  of  coplanar  forces,  it  may  be  shown 
that  only  three  independent  dynamical  equations  can  be  obtained  by 
expressing  the  equivalence  of  the  effective  forces  and  the  external 
forces.  The  remarks  made  in  Art.  387  as  to  the  meaning  of  D'Alem- 
bert's principle  must  here  be  kept  in  mind.  The  statement  that  the 
system  of  effective  forces  is  equivalent  to  the  system  of  external 
forces  is  merely  a  concise  expression  of  the  fact  that  the  two  sys- 
tems satisfy  exactly  the  same  conditions  which  are  satisfied  by  two 
systems  of  external  forces  which  would  be  equivalent  in  effect  if  ap- 
plied to  a  rigid  body.  This  relation  of  equivalence  between  the  two 
systems  is  completely  expressed  by  three  equations  falling  under  the 
above  general  forms  {a)  and  (b). 

The  three  independent  equations  may  therefore  be  written  in 
three  different  ways,  as  follows  : 

(1)  By  resolving  in  each  of  two  directions  and  taking  moments 
about  any  axis. 

(2)  By  taking  moments  about  each  of  two  axes  and  resolving  in 
a  direction  not  perpendicular  to  the  plane  containing  the  axes. 


378  THEORETICAL    MECHANICS. 

(3)  By  taking  moments  about  each  of  three  axes  not  lying  in 
one  plane.* 

449.  Equation  of  Moments  for  Axis  Not  Containing  Mass- 
Center. —  The  simplest  moment  equation,  so  far  as  regards  the  effect- 
ive forces,  is  obtained  by  taking  the  axis  through  the  mass-center  ; 
the  translational  effective  forces  are  thus  eliminated,  since  their  re- 
sultant is  applied  at  the  mass-center.  But  it  may  be  advantageous 
to  take  another  moment  axis  for  the  purpose  of  eliminating  certain 
of  the  external  forces.  In  writing  such  an  equation,  the  translational 
effective  force  Mp,  applied  at  the  mass-center,  must  not  be  omitted. 

To  illustrate  the  general  method  of  writing  an  equation  of  mo- 
ments, consider  the  problem  of  a  cylinder  rolling  on  an  inclined 
plane  (Art.  447,  Problem  I).  If  moments  are  taken  about  the  ele- 
ment of  contact  of  the  cylinder  and  plane,  the  forces  N  and  T  are 
eliminated.  The  only  external  force  whose  moment  is  not  zero  is 
the  weight  of  the  body,  and  its  moment  is  Mga  sin  /3.  The  sum  of 
the  moments  of  the  effective  forces  is 

Ma(d2x/dt'z)  +  Mk2(d20/dt2)  =  (3^72)  (;/20/^2). 

The  equation  of  moments  is  therefore 

^Ma2/2)(d20/dt2)  =  Mga  sin  0, 

which  agrees  with  the  result  given  in  Art.  447. 

Examples. 

1.  A  uniform  bar  is  placed  in  a  smooth  hemispherical  bowl  in 
such  a  position  that  a  vertical  plane  through  the  axis  of  the  bar 
contains  the  center  of  the  sphere.  Determine  a  differential  equation 
for  the  motion.  If  a  is  the  length  of  the  bar  and  c  its  distance  from 
the  center  of  the  sphere,  show  that  its  motion  will  be  the  same  as 
that  of  a  simple  pendulum  of  length  c  -J-  a'2/ 12c.  Interpret  the  lim- 
iting cases  c  =  o,  a  =  o. 

2.  A  straight  bar  is  placed  at  rest  with  one  end  on  a  smooth  hor- 
izontal plane  and  the  other  against  a  smooth  vertical  plane  which  is 
at  right  angles  to  the  vertical  plane  containing  the  axis  of  the  bar. 
Write  three  independent  equations  for  the  motion,  one  of  which 
shall  be  independent  of  the  unknown  reactions. 

*  It  is  of  course  understood  that  the  axes  of  moments  are  all  perpen- 
dicular to  the  plane  of  the  motion,  and  the  directions  of  resolution  parallel 
to  that  plane. 


PLANE    MOTION   OF    A    RIGID    BODY.  379 


§  4.   Statics  of  a  Rigid  Body. 

450.  Balanced  Forces.— A  set  of  external  forces  which,  applied 
to  a  rigid  body,  produces  no  effect  upon  its  motion,  constitutes  a 
balanced  system  or  a  system  in  equilibrium  (Art.  57).  The  condi- 
tions which  such  a  system  must  satisfy  have  been  considered  in  Part  I 
under  the  head  of  Statics.  It  may  now  be  shown  that  the  conditions 
of  equilibrium  are  deducible  from  the  general  equations  of  motion  of 
a  rigid  body. 

451.  Equilibrium  of  Rigid  Body. —  If  all  the  external  forces 
acting  upon  a  rigid  body  constitute  a  balanced  system,  the  body  is 
said  to  be  in  equilibrium. 

Equilibrium  is  not  synonymous  with  rest  ;  but  a  body  in  equilib- 
rium can  have  only  such  motion  as  it  could  have  if  acted  upon  by  no 
external  force. 

A  single  particle  acted  upon  by  no  force  or  by  balanced  forces  is 
either  at  rest  or  moving  uniformly  in  a  straight  line.  The  possible 
motion  of  a  rigid  body  acted  upon  by  balanced  forces  will  be  consid- 
ered below.  It  will  be  shown  that  the  individual  particles  of  such  a 
body  are  not  necessarily  in  equilibrium. 

452.  Equivalent  Systems  of  Forces. —  The  equations  of  motion 
of  a  rigid  body  show  that  two  systems  of  external  forces  are  equivalent 
in  their  effect  upon  the  motion  if  they  satisfy  the  following  conditions : 

(1)  Their  resolved  parts*  in  any  direction  are  equal. 

(2)  Their  moments  about  an  axis  containing  the  mass-center  are 
equal. 

For,  the  second  members  of  equations  (1 ),  (2)  and  (3)  of  Art.  446 
obviously  have  identical  values  for  two  systems  satisfying  these  con- 
ditions. 

Equivalent  forces. — These  conditions  of  equivalence  are  satisfied 
by  two  forces  which  are  equal  in  magnitude  and  direction  and  have 
the  same  line  of  action,  even  if  applied  at  different  points  in  that  line. 
This  is  the  principle  assumed  in  Art.  82  as  one  of  the  fundamental 
laws  of  Statics. 

*By  the  "resolved  part  of  the  system"  is  meant  the  sum  of  the  resolved 
parts  of  the  several  forces ;  by  the  "moment  of  the  system "  is  meant  the  sum 
of  the  moments  of  the  several  forces. 


380  THEORETICAL    MECHANICS. 

453.  Resultant  Force  or  Resultant  Couple.—  A  system  of 
coplanar  forces  is  in  general  equivalent  to  a  single  force.  For,  what- 
ever may  be  the  values  of  X,  Y  and  L  for  the  system,  a  single  force 
can  generally  be  found  having  the  same  values  of  these  three  quanti- 
ties, and  thus  giving  the  same  equations  of  motion.  Moreover,  the 
values  of  X,  Y  and  L  completely  determine  the  magnitude,  direction 
and  line  of  action  of  the  force. 

If  the  vector  sum  of  the  given  forces  is  zero,  X  and  Y  are  zero 
whatever  the  directions  of  the  axes.  In  such  a  case  there  is  no  single 
force  equivalent  to  the  given  system.  The  simplest  equivalent  system 
is  then  a  couple  whose  moment  is  equal  to  the  value  of  L  for  the 
system. 

454.  Conditions  of  Equilibrium.  —  In  order  that  a  body  acted 
upon  by  any  external  forces  may  move  as  if  acted  upon  by  no  force, 
the  values  of  X,  Y  and  L  in  the  equations  of  motion  must  be  zero. 
That  is,  the  conditions  of  equilibrium  for  a  rigid  body  are  (1)  that 
the  sum  of  the  resolved  parts  of  the  external  forces  is  zero  for  each  of 
two  rectangular  directions,  and  (2)  the  sum  of  the  moments  is  zero  for 
an  axis  containing  the  mass-center.  It  may  be  shown  as  in  Art.  104 
that  if  these  conditions  are  satisfied, 

( 1 )  The  sum  of  the  resolved  parts  in  any  direction  is  zero,  and 

(2)  The  sum  of  the  moments  about  any  axis  is  zero. 

From  these  general  principles  may  be  derived  all  the  special 
conditions  of  equilibrium  deduced  in  Chapter  V. 

455.  Motion  Under  Balanced  Forces. —  The  nature  of  the  pos- 
sible motion  of  a  rigid  body  when  the  external  forces  are  balanced 
may  be  seen  from  equations  (1),  (2)  and  (3),  Art.  446.    These  become 

d*x/dt*  =  o,  .  .         .     (1) 

dy/dt*  =  o,       .      .      .      .    (2) 

d20ldt2  =  o (3) 

Equations  (1)  and  (2)  express  the  fact  that  the  acceleration  of  the 
mass-center  is  zero,  and  its  velocity  therefore  constant.  Equation  (3) 
shows  that  the  angular  acceleration  is  zero  and  the  angular  velocity 
therefore  constant. 

It  is  thus  seen  that  if  every  particle  of  the  body  is  initially  at  rest, 
it  will  remain  at  rest.  If  initially  there  is  a  motion  of  translation,  this 
motion  will  continue  with  unvarying  velocity.      But  in  general  the 


PLANE    MOTION   OF    A    RIGID    BODY.  381 

motion  will  be  composed  of  a  translation  in  which  the  mass-center 
moves  with  unvarying  velocity,  together  with  a  rotation  with  uniform 
angular  velocity. 

The  motion  of  any  individual  particle  is  the  resultant  of  the  com- 
ponents due  to  the  translation  and  to  the  rotation.  In  general  the 
velocity  of  a  particle  not  in  a  line  perpendicular  to  the  plane  of 
the  motion  and  containing  the  center  of  mass  is  variable  both  in 
magnitude  and  in  direction.  Its  two  components  of  acceleration  may 
be  determined  as  in  Arts.  441,  443.  With  the  notation  there  used, 
the  translational  component/  is  zero.  Of  the  two  components  due 
to  the  rotation  the  one  perpendicular  to  r  is  zero,  because  the  angu- 
lar acceleration  is  zero ;  the  other  component  is  r<o2  directed  toward 
the  axis  containing  the  mass-center.  This  component  is  therefore 
the  resultant  acceleration  of  the  particle.  This  resultant  reduces  to 
zero  only  for  particles  lying  in  an  axis  containing  the  center  of  mass. 
Except  in  the  special  case  of  translation,  therefore,  the  individual 
particles  not  lying  in  this  axis  are  not  in  equilibrium. 

The  acceleration  of  any  particle  is  determined  by  the  resultant 
of  the  forces  exerted  upon  it  by  other  particles  of  the  body.  These 
internal  forces,  taken  as  a  whole  throughout  the  body,  consist  of 
stresses  (Art.  36),  so  that  the  entire  system  of  internal  forces  satisfies 
the  same  conditions  which  are  satisfied  by  a  system  of  balanced  ex- 
ternal forces.  But  the  forces  acting  upon  every  individual  particle 
are  not  balanced,  except  in  the  very  special  case  of  uniform  trans- 
latory  motion. 


CHAPTER   XXII. 

THE    PRINCIPLE   OF    IMPULSE    AND    MOMENTUM. 

§  i.  Any  System  of  Particles. 

456.  Momentum  of  a  System.— The  momentum  of  a  particle 
has  been  defined  as  a  vector  quantity  equal  to  the  product  of  the 
mass  into  the  velocity.  The  vector  sum  of  the  momenta  of  all  the 
particles  of  a  system  may  be  called  the  momentum  of  the  system. 

Restricting  the  discussion  to  plane  motion,  and  specifying  the 
position  of  every  particle  by  its  coordinates  referred  to  fixed  rectan- 
gular axes,  the  axial  components  of  the  momentum  of  a  system  are 

mxxx  +  mtxt  +     .     .     . 

and  m1  yx  -\-  m2y2  -f-     .     .     .     . 

But  if  x, y  are  the  coordinates  of  the  mass-center  and  M  is  the  whole 
mass  mx  -\-  m2  -\-     .     .     .     ,  we  have  as  in  Art.  377, 

mxxx  -f  m2x2  +     .     .     .     =  M(dx/dt); 

mxyx  -f-  m2y2  +...==  M(dy/dt). 

The  total  momentum  of  the  system  has  therefore  the  same  value  as 
if  the  entire  mass  were  moving  with  the  mass-center. 

457.  Angular  Momentum  of  a  System. —  The  motion  of  a  par- 
ticle is  at  every  instant  directed  along  a  definite  right  line  in  space. 
Momentum  may  therefore  be  regarded  as  a  localized  vector  quantity 
of  which  this  line  is  the  position- line.  The  momentum  of  a  particle 
thus  has  a  definite  moment  about  any  origin  in  the  plane  of  the  mo- 
tion. The  moment  of  the  momentum  of  a  particle  about  any  point  is 
also  called  its  angular  momentum  about  that  point  (Art.  327). 

By  the  angular  momentum  of  a  system  of  particles  about  any 
point  is  meant  the  sum  of  the  angular  momenta  of  the  individual  par- 
ticles about  that  point. 

If  the  origin  of  coordinates  be  taken  as  origin  of  moments,  the 
angular  momentum  of  a  particle  whose  mass  is  mx  and  whose  coor- 
dinates are  xx ,  yx  is 

^xxyx—  yxxx)\ 


THE    PRINCIPLE   OF    IMPULSE    AND    MOMENTUM.  383 

and  the  angular  momentum  of  the  system  is 

H  =  m^{xxyx  —  yxxx)  +  m2(x.J2  —  y2x2)  +     .     .     . 
=  Hm{xy  — yx). 

458.  External  Impulse. —  The  impulse  of  a  force  has  been  de- 
fined in  Arts.  315,  316  and  330.  Impulse  may  be  described  briefly 
as  the  time-integral  of  a  force.  It  is  a  vector  quantity,  whose  direc- 
tion coincides  with  that  of  the  force  during  every  elementary  interval 
of  time.  Any  variation  in  the  direction  of  the  force  must  be  taken 
into  account  in  the  integration. 

The  impulse  of  a  set  of  forces  may  be  defined  as  the  vector  sum 
of  their  several  impulses. 

The  vector  sum  of  the  impulses  of  all  the  external  forces  acting 
upon  the  particles  of  a  system  may  be  called  the  external  impulse  for 
the  system. 

If  X  is  the  sum  of  the  ^-components  and  Fthe  sum  of  the  j-com- 
ponents  of  all  the  external  forces,  X  and  Y  are  also  the  axial  compo- 
nents of  the  vector  sum  of  the  external  forces.  The  axial  components 
of  the  external  impulse  during  the  interval  from  t  =  t'  to  /  =  t"  are 

therefore  rt"  M" 

Xdt    and  Ydt. 

J  v  J  v 

459.  Angular  Impulse. —  Since  a  force  has  at  every  instant  a 
definite  line  of  action  in  space,  its  impulse  during  an  elementary  time 
may  be  regarded  as  a  localized  vector  quantity  whose  position-line  is 
determined  by  the  line  of  action  of  the  force.  Hence  the  impulse  has 
a  definite  moment  about  any  origin.  The  moment  of  the  impulse  for 
any  time  during  which  the  line  of  action  of  the  force  varies  is  to  be 
found  by  adding  the  moments  of  the  impulses  for  the  elementary 
intervals. 

If  L  is  the  moment  of  a  force  about  any  point,  its  angular  impulse 
for  the  interval  from  f  =  /'  to  tf  ==  /*  is 

*t* 

Ldt. 
f 

For  L  dt  is  equal  to  the  moment  of  the  impulse  of  the  force  for  the 

elementary  time  dt.      (Art.  335.) 

The  angular  impulse  of  a  set  of  forces  may  be  defined  as  the 

sum  of  the  angular  impulses  of  the  individual  forces.     If  L  denotes 

the  sum  of  the  moments  of  the  forces, 


384  THEORETICAL    MECHANICS. 

Ldt 

is  the  sum  of  the  moments  of  their  impulses,  or  their  total  angular  im- 
pulse. 

The  sum  of  the  angular  impulses  of  all  external  forces  acting  upon 
a  system  of  particles  may  be  called  the  external  angular  impulse  for 
the  system.     Its  value  is 


f. 


t" 

Ldt 
f 


if  L  is  the  sum  of  the  moments  of  the  external  forces. 

460.  Time-Integrals    of   General   Equations    of    Motion. —  In 

Chapter  XVIII  were  deduced  three  general  equations  for  the  plane 
motion  of  a  system  of  particles.  Two  of  these  (Art.  380)  are  equa- 
tions of  motion  of  the  mass-center,  and  one  (Art.  383)  was  called  the 
equation  of  angular  motion.  Using  the  same  notation  as  in  Chapter 
XVIII,  except  that  the  total  mass  is  represented  by  M and  the  coor- 
dinates of  the  mass-center  by  x,  y,  the  three  equations  are 

X =  M{d2xldfl);  .  .  .  .  (1) 
Y=M(dy/dt2);  .  .  .  .  (2) 
L  =  dHjdt.  ....     (3) 

The  first  members  of  these  equations  depend  only  upon  the  ex- 
ternal forces  ;  X  is  the  sum  of  their  jr-components,  Y  the  sum  of 
their  /-components,  L  the  sum  of  their  moments  about  the  origin  of 
coordinates.  The  quantity  designated  by  H  is  the  angular  mo- 
mentum of  the  system  of  particles  about  the  origin  of  coordinates. 

Let  each  equation  be  integrated  with  respect  to  the  time  between 
limits  t  =  t'  and  /  =  t".  Let  x\  x"  be  initial  and  final  values  of  x 
or  dx/dt,  with  similar  notation  for  y  and  H.     Then 

C  Xdt  =  M(x"  —  x')\      .         .         .     (4) 

*'  t' 

C'Ydt=M{y"-y'y,      ...     (5) 


f 


L  dt  =  H"  —  H'.  ...     (6) 


The  first  member  of  equation  (4)  represents  the  ^-component  of 
the  total  external  impulse.     The  second  member  is  the  increment  of 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  385 

the  ^r-component  of  the  total  momentum.  The  members  of  (5)  have 
similar  meanings. 

The  first  member  of  (6)  is  the  external  angular  impulse.  The 
second  member  is  the  increment  of  the  angular  momentum. 

Since  a  vector  quantity  is  completely  determined  when  its  axial 
components  are  known,  equations  (4)  and  (5)  express  the  following 
proposition : 

(1)  The  change  in  the  momentum  of  a  system  during  any  inter- 
val of  time  is  equal  in  magnitude  and  direction  to  the  impulse  of  the 
external  forces  during  that  interval. 

Equation  (6)  may  be  expressed  in  words  as  follows : 

(2)  The  change  in  the  angular  momentum  of  a  system  during 
any  interval  of  time  is  equal  to  the  angular  impulse  of  the  external 
forces  for  that  interval. 

Together  these  two  propositions  express  the  general  principle  of 
the  equivalence  of  impulse  and  momentum-increment  for  any  system 
of  particles. 

461.  Sudden  Impulse. —  Equations  (4),  (5)  and  (6)  are  wholly 
general,  and  may  be  applied  to  the  solution  of  any  problem  in  which 
the  expressed  integrations  can  be  effected.  They  are,  however,  more 
especially  useful  when  the  impulses  are  sudden  (Arts.  319-321).  If 
a  very  great  force  acts  for  a  very  brief  time,  the  positions  of  the  par- 
ticles change  very  slightly  during  that  time.  If  the  changes  of 
position  be  assumed  zero,  the  application  of  the  equations  to  partic- 
ular problems  is  simplified.  In  considering  the  effect  of  a  blow,  or 
the  impact  of  one  body  against  another,  the  impulse  is  usually  treated 
as  if  strictly  instantaneous  ;  for  most  purposes  the  results  are  suffi- 
ciently near  the  truth. 

In  dealing  with  sudden  impulses,  the  value  of  the  force  is  usually 
wholly  unknown.  The  value  of  the  impulse  cannot  therefore  be 
determined  by  integration,  but  can  be  known  only  from  its  effect. 
For  convenience,  the  integral  expressions  in  equations  (4),  (5)  and 
(6)  may  be  replaced  by  single  symbols.  The  equations  may  be 
written  in  the  form 

X-  =  A(Mx);       .         .         .         .     (7) 

r  =  A(J//»;       .         .         .         .     (8) 

L'=Aji.  .        .        .        .    (9) 

25 


386  THEORETICAL    MECHANICS. 

Here  X'  and  Y'  are  the  axial  components  of  the  total  impulse,  L  is 
the  total  angular  impulse  : 

X'=f  Xdt;     Y'  =  f    Ydt\    L'  =  C   Ldt. 

The  symbol  A  denotes  an  increment;  A(Mx)  and  A(Mj>)  are  the 
axial  components  of  the  increment  of  momentum,  and  AH  is  the 
increment  of  the  angular  momentum. 

If  the  line  of  action  of  an  instantaneous  impulse  is  known,  L  can 
be  expressed  in  terms  of  X'  and  Y'. 

462.  Linear  Momentum.— It  has  been  shown  that  the  axial 
components  of  the  total  momentum  have  the  same  values  as  if  the 
entire  mass  were  moving  with  the  mass-center.  This  is  true  for  any 
system  of  particles,  whether  rigid  or  not.  The  effect  of  an  impulse 
on  the  motion  of  the  mass-center  may  therefore  be  determined  as  if 
the  system  were  a  single  particle.  The  momentum,  computed  as  if 
the  whole  mass  were  concentrated  at  the  mass-center,  may  be  called 
the  linear  momentum  of  the  system. 

The  angular  momentum  has  not  in  general  the  same  value  as 
if  the  mass  were  concentrated  at  the  mass -center. 

Examples. 

1.  A  body  of  mass  20  lbs.  falls  from  rest  under  the  action  of 
gravity.  Determine,  by  the  principle  of  impulse  and  momentum, 
the  velocity  of  the  mass-center  after  5  sec. 

2.  A  body  whose  shape  is  that  of  a  right  circular  cylinder,  resting 
with  one  end  upon  a  smooth  horizontal  plane,  is  set  in  motion  by  the 
tension  in  a  flexible  cord  unwinding  from  its  surface.  If  the  mass  of 
the  body  is  12  lbs.  and  the  tension  is  constantly  equal  to  2  lbs. 
weight,  what  will  be  the  velocity  of  the  mass-center  after  5  sec? 
Does  the  result  depend  upon  the  position  of  the  mass-center  in  the 
body? 

3.  A  body  of  40  lbs.  mass  and  of  any  shape,  initially  at  rest, 
receives  a  blow  equivalent  to  a  force  of  100  lbs.  acting  for  y2  sec. 
What  can  be  determined  as  to  the  motion  of  the  mass-center  imme- 
diately after  the  blow? 

4.  Two  bodies  of  masses  10  lbs.  and  18  lbs.  respectively  are  con- 
nected by  a  string.  Both  being  initially  at  rest,  the  mass  of  10  lbs. 
receives  an  impulse  equivalent  to  a  force  of  50  lbs.  acting  for  1  sec. 
(a)  What  is  the  velocity  of  the  mass-center  of  the  system  immediately 
after  the  impulse  ?     (b)  If,  immediately  after  the  impulse,  the  mass 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  387 

of  18  lbs.  is  at  rest,  what  is  the  velocity  of  the  mass-center  of  the 
other  body  ?  (V)  If,  at  a  certain  instant  after  the  impulse,  the  mass- 
center  of  the  mass  of  18  lbs.  has  a  velocity  of  10  ft.-per-sec.  in  the 
direction  opposite  to  that  of  the  impulse,  what  velocity  has  the  mass- 
center  of  the  other  body  ?     Ans.  (c)  18  -f-  5g  =  179 ft.-per-sec. 

5.  Two  particles  whose  masses  are  5  lbs.  and  1  lb.  respectively, 
connected  by  an  inextensible  weightless  cord  10  ft.  long,  are  initially 
at  the  same  point.  The  lighter  body  is  projected  vertically  upward 
with  a  velocity  of  40  ft.  -per-sec.  (a)  What  will  be  the  velocity  of 
their  common  mass-center  immediately  after  the  second  particle  begins 
to  rise?     (<£)  How  long  will  the  mass-center  continue  to  rise? 

An s.  (a)  5. 1 5  ft. -per-sec.     (£)  0.16  sec. 


§  2.  Rigid  Body  Having  Motion  of  Translation  or  of  Rotation 
About  a  Fixed  Axis. 

463.  Angular    Momentum    in    Case    of   Translation. —  If  the 

motion  of  a  body  is  a  translation,  the  motion  of  every  particle  is 
known  when  that  of  the  mass-center  is  known.  The  equation  of 
angular  impulse  and  angular  momentum  therefore  is  not  needed  for 
the  determination  of  the  motion.  The  value  of  the  angular  mo- 
mentum in  case  of  translation  will,  however,  be  found  useful. 

If  v  is  the  translational  velocity,  the  momenta  of  the  several  par- 
ticles are  parallel  and  equal  to  mxvy  m2vy  ....  These  momenta 
form  a  system  analagous  to  a  system  of  parallel  forces  applied  to 
the  particles  and  proportional  to  their  masses.  The  sum  of  their 
moments  is  therefore  the  same  as  that  of  a  momentum  Mv  at  the 
mass-center.  That  is,  in  computing  the  angular  momentum  of  a 
body  whose  motion  is  translatory,  the  whole  mass  may  be  assumed 
to  be  concentrated  at  the  center  of  mass. 

464.  Linear  Momentum  of  Rotating  Body. —  Let  a  rigid  body 
of  mass  M  be  rotating,  with  angular  velocity  co,  about  a  fixed  axis 
whose  distance  from  the  center  of  mass 

is   a.     Let   A    (Fig.    183)    be   the   mass-  WM>U> 

center  and  O  the  projection  of  the  fixed  \M      yVzaccf 

axis.       The   velocity   of    the    mass-center 

is   aco   perpendicular   to    OA  ;    hence   the 

linear  momentum   is  Maco   perpendicular     & 

to  OA.  Fig.  183. 


388  THEORETICAL    MECHANICS. 

465.  Angular  Momentum  of  Rotating  Body. —  The  angular 
momentum  about  the  fixed  axis  of  rotation  may  be  computed  as  fol- 
lows : 

Let  B  (Fig.  183)  be  the  position  of  a  particle  of  mass  mx  distant 
rx  from  the  axis  of  rotation.  Its  momentum  is  m^co  perpendicular 
to  OB,  and  the  moment  of  this  momentum  about  O  is  mxrx(o.  The 
total  moment  of  momentum  about  O  is  therefore 

H  ==  {mxrx   -j-  m2r*  +     •     •     •     )(o  =  Iw, 

if  /  is  the  moment  of  inertia  of  the  body  with  respect  to  the  axis  of 
rotation. 

466.  External  Impulse  Acting  on  Rotating  Body. —  If  a  body 
is  constrained  to  rotate  about  a  fixed  axis,  it  is  convenient  to  classify 
the  external  forces  into  applied  forces  and  constraining  forces ;  the 
latter  being  usually  wholly  unknown,  except  as  the  solution  of  the 
equations  of  motion  serves  to  determine  them.  Similarly,  the  total 
external  impulse  is  made  up  of  the  applied  impulse  and  the  con- 
straining impulse.  The  latter  is  the  impulse  of  the  hinge-reaction 
(Art.  422);  it  is  unknown  in  magnitude  and  direction,  but  it  acts 
through  the  axis  of  rotation. 

467.  Equations  of  Impulse  and  Momentum  for  Rotating  Body 
Acted  Upon  by  Instantaneous  Impulse. —  Let  momentum  and  im- 
pulse be  resolved  along  the  line  joining  the  mass-center  A  to  the 
center  of  rotation  O  (Fig.  184).  Let  P'  be  the  applied  impulse,  and 
R '  the  impulse  of  the  hinge-reaction.  Let  the  components  of  P'  be 
P'n  in  direction  A  O  and  Pt  perpendicular  to  A  O ;  and  let  the  cor- 
responding  components   of  R'   be  N' 

fT'  and  T'.     (Fig.  184  represents  P'n  and 

<«^^^  /  V     P't  as  if  applied  at  A  ;   this  is  not  nec- 

**         O" "'---.._#  /       essarily    true,     but    the     figure    shows 

P^*^/         correctly  the  directions  of  the  compo- 
A. 

nent  impulses.) 

Fig.  184.  Assuming  the  impulse  to  be  instan- 

taneous, the  position  of  the  mass-center 
A  is  the  same  immediately  before  and  immediately  after  the  impulse. 
The  direction  of  motion  of  A  is  therefore  not  changed  by  the  im- 
pulse, but  its  velocity  is  changed  by  aAco  if  A<w  is  the  increment  of 
the  angular  velocity.     The  change  of  the  linear  momentum  there- 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  389 

fore  has  the  value  Ma  Aco  directed  at  right  angles  to  OA  ;  and  the 
linear  equations  of  impulse  and  momentum  take  the  form 

P;+  T  ==  Ma  Aco;           .          .         .     (1) 
P'n  +  N'=o (2) 

If  L  is  the  moment  of  the  external  impulse  and  /  the  moment 
of  inertia  of  the  body,  both  taken  about  the  axis  of  rotation,  the 
equation  of  angular  impulse  and  angular  momentum  takes  the  form 

L'  =  /Ao,.  .         .         .         .     (3) 

And  since  the  moment  of  the  constraining  impulse  R '  is  zero,  L '  is 
equal  to  the  moment  of  the  applied  impulse  (Art.  466). 

468.  Equations  of  Impulse  and  Momentum  for  Rotating  Body 
When  Impulse  Is  Not  Sudden. —  Equations  (1)  and  (2)  are  not 

valid  unless  the  impulse  is  instantaneous  ;  for  otherwise  the  direction 
of  motion  of  the  mass-center  changes  during  the  impulse,  and  the 
change  in  the  linear  momentum  has  not  the  value  used  in  these 
equations. 

Equation  (3),  however,  holds  good  whatever  the  time  occupied 
by  the  impulse  ;  for  the  increment  of  the  angular  momentum  is  always 
equal  to  I  Aco,  if  Aco  is  the  total  increment  of  the  angular  velocity.  In 
applying  the  equation  it  may  be  necessary  to  determine  L  by  inte- 
gration. 

Examples. 

1.  A  uniform  straight  bar,  hinged  at  one  end,  receives  a  known 
sudden  impulse  at  a  certain  point,  in  a  direction  perpendicular  to  the 
length.  Required  to  determine  (a)  the  ef- 
fect on  the  motion  and  (b)  the  constraining     -g. 


impulse.                                                                   j__           . 
Let  AB  (Fig.  185)  be  the  bar,  hinged      i_ a 

at  A,  and  let  C  be  the  point  of  application 

of  the  impulse.  FlG-  l85- 

Let  M  =  mass  of  bar;  a  =  its  length; 
k  =  radius  of  gyration  about  axis  of  rotation  ;   {k2  =  a2/ 3  ;)  P'  = 
magnitude  of  applied  impulse;    b  =  distance   AC;    ft)' =  angular 
velocity  just  before  the  impulse  ;  co"  =  angular  velocity  just  after  the 
impulse ;  Aco  =  co"  —  co'  =  increment  of  angular  velocity. 

(a)  Equation  (3)  becomes 

P'b  =  Mk2Aco, 
whence  Aco  =  P'b/Mk2  =  ^Pd/Ma* ; 

from  which  co"  can  be  determined  if  co'  is  known. 


390  THEORETICAL    MECHANICS. 

(fr)  Since  P't  =  P'  and  P'n  =  o,  equations  (i)  and  (2)  give 
T  =  iMaA<o  —  P'  =  (3^/2^  —  i)P'; 
N'  =  0. 
The  resultant  hinge-impulse  is  therefore 

R'  =  P'(3^  —  2a)l2at 
its  direction  being  that  of  P' . 

2.  Solve  Ex.  1  with  the  following  data :  Mass  of  bar  =  20  lbs. ; 
arm  of  impulse  =  2  ft. ;  length  of  bar  =  4  ft. ;  impulse  equivalent  to 
a  force  of  100  lbs.  acting  for  y2  sec. ;  initial  angular  velocity  =  o. 

An s.  Aft)  =  15^/16  rad. -per-sec.      T'  =  —  P/4-. 

3.  At  what  point  must  a  uniform  bar,  hinged  at  one  end,  be 
struck  in  order  that  the  hinge  may  receive  no  shock  ? 

Ans.  With  notation  of  Ex.  1,  b  =  2^/3. 

4.  If  a  bar  of  length  3  ft.  and  mass  24  lbs.  is  hinged  at  the  middle 
point  and  struck  at  a  point  1  ft.  from  the  end  with  an  impulse  equiv- 
alent to  that  of  a  force  of  1,000  lbs.  acting  for  o.  1  sec,  determine  (a) 
the  motion  just  after  the  impulse  and  (b)  the  constraining  impulse. 

Ans.  (a)  ft)"  =  25^/9  rad. -per-sec.  {b)  T'  =  — P'  =  —  loog 
poundal-sec. 

5.  A  homogeneous  circular  cylinder  is  free  to  rotate  about  its 
geometrical  axis.  A  string  wrapped  about  the  cylinder  is  jerked  in 
a  direction  perpendicular  to  the  axis.  If  the  body  receives  a  given 
increment  of  angular  velocity,  required  the  value  of  the  impulse  ap- 
plied through  the  string,  and  of  the  impulse  of  the  hinge-reaction. 

Ans.   T'  =  —  P'  =  \Ma  Aft),  i{a  =  radius  of  circular  section. 

6.  A  homogeneous  circular  cylinder  is  free  to  rotate  about  its 
axis  of  figure,  which  is  horizontal.  Around  the  cylinder  is  wrapped 
a  string,  the  free  end  of  which  hangs  vertically  and  sustains  a  body 
of  known  mass.  This  body  is  lifted  vertically  and  then  dropped  so 
that  its  velocity  is  Fjust  as  the  string  tightens.  Immediately  after 
the  impulse  its  velocity  is  V.  Required  (a)  the  angular  velocity  of 
the  cylinder  just  after  the  impulse,  and  (b)  the  impulsive  tension  in 
the  string. 

Ans.  (a)  2m(V  —  V")/Ma  rad. -per-sec. ;  (b)  m(V—  V')',  M 
being  the  mass  of  the  cylinder,  m  the  mass  of  the  other  body,  a  the 
radius  of  the  cylinder. 

469.  Center  of  Percussion.—  If  a  body  is  free  to  rotate  about  a 
fixed  axis,  there  is  a  certain  line  along  which  an  impulse  may  be  ap- 
plied without  causing  any  hinge-impulse.  This  may  be  shown  by 
equations  (1),  (2)  and  (3)  of  Art.  467. 

Equation  (2)  shows  that  if  N'  is  to  be  zero,  P'n  must  be  zero ; 
the  direction  of  the  applied  impulse  must  therefore  be  perpendicular 
to  OA  (Fig.  184),  so  that  P'f  =  P'. 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  39 1 

Putting  T'  =  o  and  Pt  =  P'  in  equation  (i), 
P  =  MaAco. 

If  b  =  arm  of  P'  with  respect  to  the  axis  of  rotation,  equation 
(3)  (Art.  467)  becomes 

Pb  =  fAco  =  Mk2A(o. 

These  two  equations  give 

b  =  k2la. 

That   is,  the   line    of  action    of  P'    must   intersect  OA    produced 
in   a    point    0'    (Fig.    186)    such    that 

00'  =  k2/a. 

This  point  is  called  the  center  of  per- 
cussion. 

Referring  to  Art.  425,  it  will  be  seen 
that  the  center  of  percussion  coincides 
with   the   center   of  oscillation   of  the  Fig.  186. 

body  if  it  rotates  as  a  compound  pendu- 
lum, the  fixed  axis  being  horizontal  and  the  applied  force  being  the 
weight  of  the  body.* 

Examples. 

1 .  A  square  plate  of  uniform  small  thickness  and  uniform  density 
is  free  to  rotate  about  one  edge  as  a  fixed  axis.  At  what  point  must 
it  be  struck  in  order  that  the  hinge-impulse  shall  be  zero  ? 

Ans.  At  a  distance  from  the  axis  equal  to  two-thirds  the  length 
of  a  side  of  the  square. 

2.  The  body  described  in  Ex.  1  is  free  to  rotate  about  an  axis 
passing  through  one  vertex  of  the  square  and  perpendicular  to  its 
plane.     Where  is  the  center  of  percussion  ? 

Ans.  At  a  distance  from  the  axis  equal  to  two- thirds  the  diagonal. 

3.  A  body  is  free  to  rotate  about  a  fixed  axis  passing  through 
the  center  of  mass.  Show  that  the  center  of  percussion  is  at  an  in- 
finite distance  from  the  axis. 

*  The  above  discussion  assumes  that  the  mass  is  symmetrically  distributed 
with  respect  to  a  plane  through  the  mass-center  perpendicular  to  the  fixed 
axis,  and  that  the  impulse  is  applied  in  this  plane  of  symmetry.  If  this  con- 
dition is  not  satisfied  the  impulse  will  (unless  certain  very  special  conditions 
are  satisfied)  tend  to  cause  rotation  about  some  axis  inclined  to  the  fixed  axis, 
and  to  resist  this  tendency  impulsive  reactions  will  act  at  the  hinge.  See 
Art,  427. 


392  THEORETICAL    MECHANICS. 

4.  A  body  free  to  rotate  about  a  fixed  axis  passing  through  the 
mass-center  receives  at  the  same  instant  two  impulses  which  are 
equal  and  opposite  but  not  collinear.  Show  that  the  hinge-impulse 
is  zero. 

5.  A  body  free  to  rotate  about  any  fixed  axis  is  acted  upon  by  an 
impulsive  couple  (i.  e. ,  two  impulses  as  in  Ex.  4)  whose  moment  is 
Q.     Determine  the  hinge-impulse. 

Ans.  With  notation  of  Art.  467,  N'  =  0,  T  =  MaQ/L 


§  3.   Resultant  Momentum. 

470.  General  Equation  of  Impulse  and  Momentum. —  Before 
taking  up  the  general  case  of  plane  motion  of  a  rigid  body,  it  will  be 
well  to  consider  the  full  meaning  of  the  general  principle  of  the 
equivalejice  of  impulse  and  momentum,  stated  in  Art.  460. 

It  was  there  shown  that  the  total  impulse  of  the  external  forces 
during  any  time  and  the  total  change  of  momentum  during  that  time 
are  equivalent ;  that  is,  they  are  related  to  each  other  in  the  same 
way  as  two  sets  of  forces,  which,  applied  to  a  rigid  body,  are  equiva- 
lent in  effect.  The  discussion  of  Art.  460  was  restricted  to  plane 
motion,  and  the  principle  was  proved  by  deducing  three  algebraic 
equations  which  express  the  equivalence  of  two  sets  of  coplanar 
forces.  It  will  be  instructive  to  give  another  deduction  which  is 
independent  of  any  particular  system  of  coordinates,  and  has  the 
further  advantage  of  showing  the  full  generality  of  the  principle. 
Starting  with  the  principle  of  equivalence  of  impulse  and  momentum- 
increment  for  a  single  particle,  it  may  readily  be  extended  to  any 
system  of  particles. 

For  a  single  particle,  the  resultant  imptilse  for  a  time  dt  is  equiv- 
alent to  the  change  of  momentum  during  that  time.  "Resultant 
impulse ' '  means  the  impulse  of  the  resultant  of  all  forces,  external 
and  internal,  which  act  upon  the  particle.  "Equivalent"  means 
equal  in  magnitude  and  direction  and  having  the  same  position-line. 
If  the  impulses  for  all  elementary  times  throughout  a  given  finite  in- 
terval be  combined  as  if  they  were  forces,  due  account  being  taken 
of  magnitude,  direction  and  position-line,  and  if  the  elementary 
momentum-increments  be  combined  in  like  manner,  the  two  result- 
ants must  be  equal. 

This  principle  may  be  applied  to  every  individual  particle  of  a 
system.     Combining  the  resulting  equations  for  all  particles,  it  fol- 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  393 

lows  that  the  resultant  impulse  for  the  system  is  equivalent  to  the 
resultant  of  all  the  momentum-increments. 

''Resultant  impulse"  here  means  the  resultant  of  the  impulses 
of  all  external  and  internal  forces  acting  upon  the  particles ;  but  it 
may  readily  be  seen  that  the  resultant  impulse  of  the  internal  forces 
is  zero.  This  follows  from  the  law  of  action  and  reaction.  The  two 
forces  exerted  by  any  two  particles  upon  each  other  are  at  every  in- 
stant equal,  opposite  and  collinear  ;  therefore  their  impulses  during 
any  time  are  equal,  opposite  and  collinear.  The  impulses  of  tlie 
internal  forces  for  tlie  entire  system  have  therefore  a  zero  resultant. 
It  follows  immediately  that 

The  resultant  external  impulse  is  equal  to  the  resultant  change 
of  momentum.  * 

This  general  principle  is  closely  analagous  to  the  principle  of  the 
equivalence  of  the  external  forces  and  the  effective  forces  (Art.  387). 
The  equation  of  impulse  and  momentum  is  in  fact  the  time-integral 
of  the  equation  of  external  and  effective  forces.     Thus, 

(resultant  external  force)  t=  —  (resultant  external  impulse) ; 

dt 

(resultant  effective  force)  =  —  (resultant  momentum). 

dt 

471.  Algebraic  Equations  of  Impulse  and  Momentum. —  The 

above  general  principle  is  true  without  restriction  to  plane  motion. 
To  express  the  principle  fully  in  the  case  of  motion  in  three  dimen- 
sions requires  six  independent  equations,  just  as  six  equations  are 
required  to  determine  the  resultant  of  a  system  of  forces  in  three 
dimensions  acting  upon  a  rigid  body  (Chapter  X).  When  the  mo- 
tion is  restricted  to  a  plane,  the  principle  is  expressed  by  three  inde- 
pendent equations,  just  as  in  the  case  of  coplanar  forces  in  Statics. 
The  three  independent  equations  may  be  {a)  two  equations  of  reso- 
lution and  one  of  moments,  {If)  one  equation  of  resolution  and  two  of 
moments,  or  (c)  three  equations  of  moments.  The  restrictions  to  be 
observed  in  choosing  origin  of  moments  and  direction  of  resolution 
are  the  same  as  in  Art.  104. 

In  the  case  of  a  sudden  impulse,  the  most  useful  equations  are 
often  (7),  (8)  and  (9)  of  Art.  461.     Equations  (7)  and  (8)  are  ob- 

*The  remarks  made  in  Art.  387  regarding  the  meaning  of  the  term  M  re- 
sultant "  are  applicable  here. 


394  THEORETICAL    MECHANICS. 

tained  by  resolving  along  each  of  a  pair  of  fixed  rectangular  axes. 
In  certain  cases,  however,  simple  equations  result  by  resolving  in 
other  directions.  In  equation  (9)  the  axis  of  moments  is  any  line 
perpendicular  to  the  plane  of  the  motion.  In  order  to  express  the 
value  of  H  in  any  case  of  plane  motion,  it  is  necessary  to  consider 
the  resultant  momentum  in  the  general  case. 

472.  Value  of  Resultant  Momentum. —  The  resultant  of  a  set 
of  coplanar  forces  is  either  a  single  force,  a  couple,  or  zero.  A  simi- 
lar proposition  is  true  of  the  resultant  momentum  of  any  system  of 
particles  moving  in  a  plane. 

If  the  vector  sum  of  the  momenta  of  the  individual  particles  is 
not  zero,  its  value  gives  the  magnitude  and  direction  of  the  resultant 
momentum.  As  in  Art.  462,  it  is  seen  that  in  general  the  resultant 
momentum  has  the  same  magnitude  and  direction  as  if  the  entire 
mass  were  moving  with  the  mass-center.  Its  line  of  action  does  not 
in  general  contain  the  mass-center,  however,  but  may  be  determined 
by  a  moment  equation. 

If  the  velocity  of  the  mass-center  is  zero,  the  vector  sum  of  the 
momenta  of  the  particles  is  zero.  In  this  case  the  resultant  mo- 
mentum is  in  general  a  couple,  whose  moment  is  the  same  for  every 
axis,  and  may  therefore  be  computed  for  whatever  axis  is  most  con- 
venient. 

If  the  velocity  of  the  mass-center  is  zero  and  the  angular  mo- 
mentum is  also  zero,  the  resultant  momentum  is  zero.  In  this  case 
if  the  system  is  rigid  it  must  be  at  rest ;  but  if  not  rigid  the  indi- 
vidual particles  are  not  necessarily  at  rest. 

473.  Resultant  Momentum  in  Case  of  Translation. —  In  the 
case  of  translation,  the  resultant  momentum  is  a  single  linear  mo- 
mentum whose  value  is  in  all  respects  (magnitude,  direction  and 
position-line)  the  same  as  if  the  entire  mass  were  moving  with  the 
mass-center ;  for  the  momenta  of  the  particles  are  proportional  to 
their  masses  and  are  parallel. 

474.  Resultant  Momentum  of  Rigid  Body  Rotating  About  a 
Fixed  Axis  Containing  the  Center  of  Mass. —  If  a  rigid  body  ro- 
tates about  an  axis  containing  the  mass-center,  the  resultant  momen- 
tum is  a  couple  (Art.  472);  its  moment  is  equal  to  the  angular 
momentum  about  the  axis  of  rotation.      The  value  of  this  moment  is 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  395 

Ico,  if  /  is  the  moment  of  inertia  with  respect  to  the  axis  of  rotation 
and  co  the  angular  velocity  (Art.  465). 

In  this  case  the  angular  momentum  has  the  same  value  for  all 
axes  parallel  to  the  axis  of  rotation. 

475.  Resultant  Momentum  of  Rigid  Body  Having  Any  Plane 
Motion. —  The  resultant  momentum  in  the  general  case  of  plane 
motion  is  best  determined  by  considering  the  actual  instantaneous 
motion  to  be  made  up  of  two  components  as  in  Art.  443: 

(a)  A  rotation,  with  the  actual  angular  velocity  of  the  body,  about 
an  axis  containing  the  mass-center. 

(b)  A  translation  whose  velocity  is  equal  in  magnitude  and  direc- 
tion to  that  of  the  mass -center. 

Let  /  be  the  moment  of  inertia  with  respect  to  the  assumed  axis, 
M  the  mass,  co  the  instantaneous  angular  velocity,  and  v  the  instan- 
taneous velocity  of  the  mass-center. 

(a)  The  resultant  momentum  corresponding  to  the  rotation  is  a 
couple  whose  moment  is  Ico. 

(b)  The  resultant  momentum  corresponding  to  the  translation  is 
a  linear  momentum  Mv  whose  position-line  contains  the  mass-center. 

These  can  be  combined  into  a  single  linear  momentum  equal  to 
Mv  at  a  distance  from  the  mass-center  equal  to  Ico/Mv,  which  is, 
therefore,  the  resultant  momentum.  In  general,  however,  it  is  con- 
venient to  use  the  two  components  (a)  and  (b)  rather  than  their 
resultant. 

476.  Angular  Momentum  About  Any  Axis. — The  angular 
momentum  about  any  axis  perpendicular  to  the  plane  of  the  motion 
may  be  found  by  taking  the  sum  of  the  moments  of  the  two  com- 
ponents of  the  resultant  momentum.  The  moment  of  the  couple  of 
course  has  the  same  value  Ico  for  every  axis. 

The  angular  momentum  about  any  axis  perpendicular  to  the 
plane  of  the  motion  is  equal  to  the  angular  momentum  about  a  par- 
allel axis  containing  the  mass-center  plus  the  angular  momentum  of 
the  entire  mass  assumed  concentrated  at  the  mass -center. 


§  4.  Any  Plane  Motion  of  a  Rigid  Body. 

477.  Equations  of  Impulse  and  Momentum  in  General  Case 
of  Plane  Motion. — We  are  now  prepared  to  apply  the  equations  of 


396  THEORETICAL    MECHANICS. 

Art.  461  to  the  unrestricted  plane  motion  of  a  rigid  body.  It  must 
be  remembered,  however,  that  the  equations  may  take  other  forms, 
depending-  upon  the  choice  of  the  coordinates  of  position  and  of  the 
axis  of  moments.  The  application  of  the  equations  may  best  be  ex- 
plained by  the  solution  of  specific  problems.  For  convenience,  the 
general  equations  are  here  repeated. 

X'  =  A(Mx);  .     (1) 

Y'  =  A(My);         ....     (2) 
L'^-AH.  ....     (3) 

In  equation  (3),  moments  are  taken  about  any  axis  perpendicular 
to  the  plane  of  motion.  If  the  axis  contains  the  mass-center,  H  == 
/ft),  and  the  equation  becomes 

V  =  /Aft> (4) 

This  is  identical  in  form  with  equation  (3)  of  Art.  467,  which  ap- 
plies to  the  case  of  rotation  about  any  fixed  axis ;  but  in  that  case  / 
is  the  moment  of  inertia  about  the  axis  of  rotation. 

478.  Effect  of  Instantaneous  Impulse  on  Free  Body  Initially 
at  Rest. —  Let  a  body,  initially  at  rest  but  free  to  move  in  a  plane, 
receive  an  instantaneous  impulse  of  known  magnitude,  direction  and 
line  of  action. 

Since  the  resultant  momentum  is  zero  before  the  impulse, 

(momentum-increment)  =  (resultant  momentum  after  impulse). 

Hence  the  motion  just  after  the  impulse  must  be  such  that  the  re- 
sultant momentum  has  the  same  magnitude,  direction  and  position- 
line  as  the  impulse.  Let  P'  be  the  magnitude  of  the  impulse,  b  the 
distance  of  its  line  of  action  from  the  mass-center,  M  =  mass  of 
body,  Mk2  =  /  =  its  moment  of  inertia  with  respect  to  the  axis 
through  the  mass  center,  v  =  velocity  of  mass-center,  and  co  —  an- 
gular velocity  just  after  the  impulse.  Then  by  Art.  475  the  resultant 
momentum  is  equal  in  magnitude  to  Mv,  its  direction  is  that  of  v} 
and  its  moment  about  the  mass-center  is  Mk2co.  Hence  the  direction 
of  v  coincides  with  that  of  P',  and 

P  =  Mv,     P'b  -     Mk2co; 
or  v  =  P/M,     co  =  P'blMk1. 

Instantaneous  axis. —  Let  A  (Fig.  187)  be  the  mass-center,  and 
draw  AB  perpendicular  to  the  line  of  action  of  P.     Since  the  velocity 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  397 

of  A  after  the  impulse  is  parallel  to  P\  the  instantaneous  motion  is 
a  rotation  about  some  axis  C  intersecting  BA  produced.  Let  A  C 
=  z ;  then  v  =  zco,  or 

z  =  v/m  =  P/6. 

The  points  C  and  B  are  thus  related  in  the  same  way  as  the  centers 
of  suspension  and  oscillation  of  a  com- 
pound pendulum  (Art.  425).  r_/-~~' — *^> 1 

The  determination  of  the  instanta-      )  u^z_  j~J 

neous  axis  and  of  the  angular  velocity     (  q  4 "_l"_j      ( 

gives  a  clear  idea  of  the  state  of  motion      \  *M       \ 

immediately  after  the  impulse.     If  the  "~"^v     Js-^ 

body  is  acted  upon  by  no  forces  sub-  Fig.  187. 

sequently,  the  angular  velocity  remains 

constant,  and  the  velocity  of  the  mass-center  remains  constant  in 
magnitude  and  direction ;  the  instantaneous  axis  therefore  continues 
at  the  same  distance  from  the  mass-center,  and  its  locus  is  a  plane 
parallel  to  the  path  of  the  center  of  mass.  The  surface  traced  in 
the  body  by  the  instantaneous  axis  is  a  circular  cylinder  of  radius  z 
whose  geometrical  axis  contains  the  mass-center. 

Application  of  equations. —  To  show  the  application  of  the  equa- 
tions of  Art.  477,  let  the  axis  of  x  have  the  direction  of  the  impulse. 
Then 

X' =  P\      Y'  =  0] 

and  if  moments  are  taken  about  the  axis  through  the  mass-center, 

L  =  P'b. 

Equations  (1),  (2)  and  (4)  therefore  become 

MAx  =  P,     MAy  =  o,     Mk2Av  =  P'b. 

But  Ax  and  Ay  are  the  components  of  v,  and  since  Ay  =  o, 

v  =  Ax=  P'jM. 

Also,  Aw  becomes  in  this  case  a),  hence 

co  =  P'bjMk\ 
as  before. 

To  illustrate  still  further  the  general  method,  let  moments  be 
taken  about  a  point  in  the  line  of  action  P' .  The  initial  angular 
momentum  about  this  point  is  zero,  the  moment  of  P'  is  zero,  hence 


398  THEORETICAL   MECHANICS. 

the  angular  momentum  about  this  point  after  the  impulse  is  zero. 
But  by  Art.  476  its  value  is 

Mtfto  —  Mvb ; 
equating  this  to  zero, 

v/e>  =  &2/b, 

which  agrees  with  the  result  already  found.     This  moment  equation 
may  take  the  place  of  one  of  the  equations  before  used. 

Examples. 

1.  A  straight  bar  of  uniform  small  cross-section  and  uniform 
density,  free  to  take  up  any  plane  motion,  receives  a  blow  at  a  given 
point  directed  at  right  angles  to  its  length.  Determine  the  instan- 
taneous axis  of  rotation. 

An s.  With  the  above  notation,  z  =  I2 1 12b,  I  being  the  length  of 
the  bar. 

2.  In  Ex.  1,  let  the  mass  of  the  bar  be  20  lbs.,  its  length  2  ft., 
and  let  the  impulse  be  equivalent  to  that  of  a  force  of  1,000  lbs.- 
weight  acting  for  o.  1  sec.  If  b  =  IJ2,  determine  v  and  co  after  the 
impulse.  Ans,  v=  $g  ft.  -per-sec  ;  co  =  1  $g  rad. -per-sec. 

3.  A  straight  bar  of  uniform  cross-section  and  uniform  density,  2 
ft.  long,  receives  a  blow  in  a  direction  perpendicular  to  its  length  at 
a  point  6  ins.  from  one  end.     Determine  the  instantaneous  axis. 

Ans.  4  ins.  from  end. 

4.  In  Ex.  1,  let  the  direction  of  the  impulse  be  inclined  at  angle 
6  to  the  length  of  the  bar.     Determine  the  instantaneous  axis. 

5.  If,  in  Ex.  3,  the  impulse  is  applied  at  the  same  point  but  in  a 
direction  inclined  300  to  the  bar,  determine  the  instantaneous  axis. 

An s.  4  ins.  from  middle  point  of  bar. 

6.  A  square  plate  of  uniform  small  thickness  and  uniform  density 
is  suspended  by  strings  attached  to  two  corners  so  that  two  edges 
are  vertical.  It  receives  a  blow  perpendicular  to  its  plane,  applied  at 
the  middle  point  of  the  lowest  edge.  Determine  the  instantaneous 
axis.  Ans.  If  a  =  side  of  square,  z  =  a/6. 

7.  A  circular  plate  of  uniform  density  and  uniform  small  thick- 
ness is  suspended  by  a  string  attached  at  a  point  in  the  circumference. 
How  must  it  be  struck  in  order  that  it  shall  begin  to  rotate  about  a 
vertical  tangent  ?     Ans.  At  a  point  bisecting  a  horizontal  radius. 

8.  If  the  plate  suspended  as  in  Ex.  7  receives  a  blow  in  a  direc- 
tion perpendicular  to  its  plane  and  at  a  distance  from  the  center 
equal  to  one-third  the  radius,  determine  the  instantaneous  axis. 

Ans.  z  =  3*2/2,  if  a  =  radius. 

9.  A  homogeneous  parallelopiped  of  mass  10  kilogr.  and  di- 
mensions (in  cm.)  20  X  30  X  40,  receives  a  blow  whose  direction  is 


THE    PRINCIPLE   OF    IMPULSE   AND    MOMENTUM.  399 

• 

perpendicular  to  one  of  the  largest  faces  and  whose  point  of  application 
is  the  middle  point  of  its  shorter  edge.  Determine  the  instantaneous 
axis.  If  the  velocity  of  the  mass-center  after  the  impulse  is  1  met.  - 
per-sec. ,  determine  the  angular  velocity  and  the  value  of  the  impulse. 
Ans.  z  =  8y3  cm. ;  co  =  12  rad. -per-sec. ;  P'=  io6  dyne-seconds. 

10.  In  the  above  general  problem,  discuss  the  case  b  =  o. 

11.  Discuss  the  case  in  which  b  is  infinite  while  P'b  is  finite. 

12.  A  horizontal  blow  must  be  applied  at  what  point  of  a  billiard 
ball  in  order  that  it  shall  begin  to  roll  without  sliding  ? 

479.  Effect  of  Impulsive  Couple  on  Free  Body  Initially  at 
Rest. —  If  a  body  is  subjected  to  two  instantaneous  impulses  which 
are  equal  in  magnitude  but  opposite  in  direction  and  not  collinear, 
the  motion  of  the  body  immediately  after  the  impulse  must  be  such 
that  the  resultant  momentum  is  a  couple.  This  requires  that  the 
instantaneous  axis  shall  pass  through  the  mass-center. 

If  L  is  the  moment  of  the  impulsive  couple,  the  angular  velocity 
after  the  impulse  is 

co  =  L'IMk\ 

Examples. 

1.  A  uniform  straight  bar  of  length  /receives  simultaneously  two 
blows  applied  close  to  the  ends,  at  right  angles  to  the  length,  and  in 
opposite  directions.  Either  blow  alone,  applied  at  the  middle  point, 
would  give  the  bar  a  translational  velocity  V.  Determine  the  motion 
after  the  impulse. 

Ans.  A  rotation  about  a  central  axis,  with  angular  velocity  12  Vjl. 

2.  A  uniform  straight  bar  of  length  /  receives  simultaneously  two 
impulses  P'  and  2P'  at  right  angles  to  the  length  and  in  the  same 
direction,  the  former  applied  at  the  middle,  the  latter  at  one  end. 
The  impulse  P'  alone  would  give  the  bar  a  translational  velocity  V. 
Determine  the  instantaneous  motion. 

Ans.  A  rotation  with  angular  velocity  12  Vjl  about  an  axis  distant 
lj\  from  the  middle  point. 

3.  A  uniform  bar  2  ft.  long  whose  mass  is  10  lbs.,  constrained  to 
rotate  about  one  end,  receives  an  impulse  at  a  point  0.25  ft.  from  the 
free  end  which  gives  it  an  instantaneous  angular  velocity  of  3600  per 
sec.  An  equal  impulse  is  applied  to  an  exactly  similar  bar  which  is 
wholly  free.     Determine  the  instantaneous  motion. 

Ans.  An  angular  velocity  of  247J-/7  rad. -per-sec.  about  an  axis  5^ 
ins.  from  the  middle  point. 

4.  A  body  is  free  to  rotate  about  a  fixed  axis  through  the  mass- 
center.     Show  that  an  impulsive  couple  will  cause  no  hinge-impulse. 


400  THEORETICAL    MECHANICS. 

480.  Effect  of  Instantaneous  Impulse  on  Free  Body  Not  In- 
itially at  Rest. —  The  momentum-ijicrement  due  to  a  given  impulse 
has  the  same  value,  whatever  the  initial  motion  of  the  body.  The 
motion  immediately  after  the  impulse  is  found  by  combining  the 
motion  due  to  the  impulse  with  the  motion  previously  existing. 

Thus  if  x,  y  are  the  axial  components  of  the  velocity  of  the  mass- 
center  and  ft)  the  angular  velocity,  and  if  values  just  before  the  impulse 
are  denoted  by  a  single  accent  and  values  just  after  the  impulse  by  a 
double  accent, 

x"  =  x    -J-  Ax, 

y"  =  y>  +  A>, 

&)"   =  ft)'  -j-  Aft) ; 

in  which  Air,  Ay  and  Aft)  are  to  be  determined  from  equations  (1), 
(2)  and  (4)  of  Art.  477. 

The  velocity  of  the  mass-center  after  the  impulse  is  equal  to  the 
vector  sum  of  its  velocity  before  the  impulse  and  that  due  to  the  im- 
pulse. The  angular  velocity  after  the  impulse  is  the  algebraic  sum 
of  the  angular  velocity  before  the  impulse  and  that  due  to  the  impulse. 

Change  of  instantaneous  axis. — The  position  of  the  instantaneous 
axis  of  rotation  immediately  after  the  impulse  is  less  simply  deter- 
mined in  case  the  body  is  in  motion  before  the  impulse  than  when  it 
is  initially  at  rest.  It  may,  however,  be  determined,  by  combining 
the  change  of  motion  due  to  the  impulse  with  the  initial  motion. 

Let  the  change  of  motion  due  to  the  impulse,  computed  as  if 
the  body  were  initially  at  rest,  be  a  rotation  about  an  instantaneous 

center  C  (Fig.  188)  with  angular  veloc- 

%~^        J*C  J^\       *ty  ^  >  *et  *ke  motion  just  before  the 

f   c\    f  C"A         f  C'\     impulse  be  a  rotation  about  an  instan- 

\^^y     \ J         V_^/      taneous  center  C  with  angular  velocity 

Fig.  188.  ft>' ;    and    let    the    motion    immediately 

after  the  impulse  be  a  rotation  about 
an  instantaneous  center  C  with  angular  velocity  ft) ".  By  Art.  437 
C"  lies  upon  the  line  CC  at  such  a  point  that 

CC"  X  ft>  =  C'C"  X  (o'y 

and  ft) "  =  ft)  -f-  ft)'. 

If  ft)'  and  ft)  have  like  signs,  C"  lies  between  C  and  C;  if  they 
have  opposite  signs,  C"  lies  on  C  C  produced. 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  4OI 

If  &)'  and  ft)  are  equal  and  opposite,  C"  is  at  infinity  on  CC '.  In 
this  case  ft)"  is  zero,  and  the  resultant  motion  is  a  translation. 

Examples. 

1.  A  free  circular  disc  of  uniform  density  and  thickness  is  rotat- 
ing about  a  central  axis  perpendicular  to  its  plane  when  it  receives  a 
blow  directed  along  a  tangent.  What  is  the  angular  velocity  after 
the  impulse  if  the  velocity  of  the  mass-center  is  v"l 

Let  r  =  radius  of  disc,  ft)'  its  original  angular  velocity,  P'  the 
impulse.     From  the  given  data, 

P'  =  Mv\ 
and  the  moment  of  P'  about  the  mass-center  is 

U  =  Mv"r 
in  magnitude,  but  may  be  either  positive  or  negative.     If  positive, 
the  equation  of  moments  about  the  mass-center  is 

Mv"/r=iMr2Aco, 
whence  Aft)  =  2v"/r;     ft)"  =  ft)'  -f-  Aft)  =  »' +  2v"jr. 

If  L  is  negative,  ft)"  =  go'  —  2v"/r. 

2.  A  rotating  wheel  falls  vertically  in  its  own  plane  and  strikes  a 
horizontal  plane  which  is  perfectly  rough  and  inelastic.  Determine 
the  subsequent  motion.  ["  Perfectly  rough"  means  that  no  sliding 
occurs  ;  "perfectly  inelastic"  means  that  the  wheel  does  not  rebound 
but  remains  in  contact  with  the  plane.] 

This  problem  may  be  solved  by  a  single  equation  obtained  by 
taking  moments  about  an  axis  through  the  point  of  contact  of  the 
wheel  with  the  plane.  The  moment  of  the  impulse  about  this  axis 
is  zero,  hence  the  angular  momentum  has  the  same  value  before  and 
after  the  impulse.  Letting  v"  denote  the  velocity  of  the  mass-center 
after  the  impulse,  and  taking  other  notation  as  usual,  the  angular 
momentum  before  the  impulse  is 

Mk2co'. 
After  the  impulse  it  is 

MkW  +  Mv"ry 

if  r  is  the  radius  of  the  wheel.     Therefore 

MkW  =  Mk2co"  +  Mv"r. 

But  also,   v"  =  rco"  ;  hence 

ft)"  =  w'k2l(k2  +  r2)  ;     v"  =  rv&Kk2  +  r2). 

3.  A  circular  disc  of  uniform  thickness  and  density,  of  radius  2 
ft.  and  mass  50  lbs.,  rotating  at  the  rate  of  100  revolutions  per 
minute,  falls  in  its  own  vertical  plane  and  strikes  a  horizontal  plane 
surface.  At  the  instant  of  striking  its  center  has  a  horizontal  velocity 
of  20  ft.-per-sec.  and  a  vertical  velocity  of  30  ft.-per-sec.  If  the 
plane  is  so  rough  that  there  is  no  sliding  and  so  inelastic  that  there 

26 


402  THEORETICAL    MECHANICS. 

is  no  rebound,  determine  (a)  the  subsequent  motion,  (b)  the  value  of 
the  impulse. 

Ans.  (a)  v"  =  20.32  ft. -per-sec.  or  — 6.36  ft. -per-sec. ,  depend- 
ing upon  the  direction  of  v'.  (b)  The  horizontal  impulse  is  either 
16  poundal-seconds  or  —  1,318  poundal-seconds;  the  vertical  impulse 
is  1 ,  500  poundal-seconds. 

4.  A  homogeneous  right  circular  cylinder,  resting  with  one  end 
upon  a  smooth  horizontal  plane,  is  set  in  motion  by  a  jerk  applied  to 
the  free  end  of  a  string  wrapped  on  its  surface.  Determine  the  in- 
stantaneous axis  of  rotation. 

An s.  At  a  distance  from  the  geometrical  axis  equal  to  half  the 
radius. 

5.  In  Ex.  4,  let  the  radius  of  the  cylinder  be  0.5  met.  and  its 
mass  5  kilogr.  If  the  angular  velocity  just  after  the  impulse  is  2 
rev.  -per-sec. ,  what  is  the  value  of  the  impulse  ?  What  is  the  velocity 
of  the  mass-center? 

Ans.  Impulse  =  250,000^  dyne-sec.  Velocity  of  mass-center 
=  507T  cm. -per-sec. 

6.  A  body  has  any  plane  motion,  when  a  certain  line  in  the  body, 
perpendicular  to  the  plane  of  the  motion,  suddenly  becomes  fixed. 
Determine  the  subsequent  motion. 

The  impulse  by  whose  action  the  stated  change  in  the  motion  is 
produced  acts  in  some  line  passing  through  the  axis  which  becomes 
fixed.  The  angular  momentum  with  respect  to  this  axis  is  therefore 
unchanged.  The  motion  after  the  impulse  may  be  determined  by 
equating  the  values  of  the  angular  momentum  before  and  after  the 
impulse. 

7.  A  homogeneous  right  circular  cylinder  is  rotating  freely  about 
its  axis  of  figure  when  an  element  of  the  cylindrical  surface  suddenly 
becomes  fixed.     Determine  the  subsequent  motion. 

Let  M  be  the  mass  of  the  cylinder,  a  its  radius,  co  its  original 
angular  velocity,  co"  its  angular  velocity  after  the  impulse.  The  orig- 
inal value  of  the  angular  momentum  about  an  axis  instantaneously 
coinciding  with  an  element  of  the  surface  is  Mcfco' '  I2.  The  angular 
momentum  about  this  axis  after  the  impulse  is  3Ma2co"J2.  Hence 
co"  =  ^co'. 

8.  In  Ex.  7,  determine  the  value  of  the  impulse. 

The  linear  velocity  of  the  mass-center  changes  from  o  to  aco" ; 
hence  the  value  of  the  impulse  is  Maco",  its  direction  being  perpen- 
dicular to  the  plane  containing  the  fixed  axis  and  the  geometrical 
axis. 

9.  A  uniform  straight  bar  of  mass  10  lbs.  and  length  2  ft.,  rotat- 
ing about  its  mass-center  at  the  rate  of  5  rev. -per-sec. ,  receives  a 
blow  equivalent  to  a  force  of  100  lbs.  acting  for  0.5  sec.  at  a  point 
3  ins.  from  one  end,  in  a  direction  perpendicular  to  the  length  of 


THE    PRINCIPLE    OF    IMPULSE    AND    MOMENTUM.  403 

the  bar.     Determine  the  instantaneous  center  and  angular  velocity 
just  after  the  blow. 

Ans.   Instantaneous  center  is  7.10  ins.  from  middle  point  of  bar. 

10.  With  the  initial  motion  described  in  Ex.  9,  {a)  what  impulse 
will  reduce  the  motion  to  a  translation  of  10  ft. -per-sec.  ?  (b)  What 
impulse  will  leave  the  body  at  rest? 

Ans.  (a)  An  impulse  of  100  poundal-seconds,  whose  line  of  ac- 
tion is  ^7r  ft.  from  the  mass-center. 

11.  Show  that  an  impulsive  couple,  applied  to  a  body  rotating 
about  an  axis  containing  the  mass-center,  does  not  change  the  axis  of 
rotation. 

12.  A  body  whose  mass  is  10  kilogr.  and  least  radius  of  gyration 
15  cm.  has  at  a  certain  instant  a  translational  velocity  of  2  met. -per- 
sec.  and  an  angular  velocity  of  2  rev. -per-sec.  Two  opposite  im- 
pulses are  simultaneously  applied  in  lines  10  cm.  apart,  each  equiva- 
lent to  a  force  of  5,000  dynes  acting  for  10  sec.  Determine  the 
subsequent  motion. 

Ans.  The  linear  velocity  is  unchanged  ;  the  angular  velocity  is 
increased  or  decreased  by  2/9  rad. -per-sec 

481.  Connected  Bodies. —  The  principle  that  the  external  impulse 
is  equal  to  the  change  of  momentum  it  produces  may  be  applied  to 
any  number  of  bodies  regarded  as  forming  a  system.  If  the  bodies 
press  against  one  another,  or  are  connected  by  strings  or  by  hinges, 
the  forces  they  exert  upon  one  another  are  internal,  and  the  impulses 
of  these  forces  may  be  omitted  from  the  equations  of  impulse  and 
momentum  for  the  system.  For  plane  motion  three  independent 
equations  can  therefore  be  written  which  do  not  involve  the  impulses 
of  these  internal  forces  or  "reactions."  These  are  not  generally 
sufficient  to  determine  the  motion.  To  complete  the  solution  it  is 
necessary  to  write  equations  for  single  bodies  of  the  system.  For 
any  given  body  it  is  often  possible  to  form  one  or  more  such  equa- 
tions not  containing  the  unknown  reactions. 

The  method  of  procedure  will  be  illustrated  by  examples. 

Examples. 

1.  Two  uniform  straight  bars  AB,  BC,  connected  by  a  frictionless 
hinge  at  B,  rest  on  a  horizontal  plane  so  that  ABC  is  a  straight  line. 
The  bar  AB  receives  at  a  certain  point  a  horizontal  blow  at  right 
angles  to  its  length.     Determine  the  motion  just  after  the  impulse. 

Let  P'  denote  the  impulse,  applied  at  distance  a  from  B  (Fig. 
189).  Let  ;;/,  m'  be  the  masses  and  /,  /'  the  lengths  of  AB  and 
irrespectively.     Since  the  velocity  of  the  mass-center  of  the  system 


404  THEORETICAL    MECHANICS. 

just  after  the  impulse  will  be  parallel  to  P' ,  it  is  obvious  that  every 

point  of  both  bars  will  instantaneously  move  parallel  to  P\      Let  v 

be  the  velocity  of  the  mass-center  of  AB  and  co  its 

A  angular  velocity,  v\  co'  being  the  velocity  of  the  mass- 

,       center  of  i>  £7  and  its  angular  velocity.     Then  v*  can 

—       be  expressed  in  terms  of  v,  co,  co',  as  follows  : 

If  u  is  the  velocity  of  B, 


a 

u    =  v  —  \lco  ; 

£  v'  =  11  —  \l'co'  =  v  —  \{lco  -f-  /'ft)'). 


Equating  the  impulse  P  to  the  linear  momentum 
„  of  the  system  after  the  impulse, 

Fig.  189.  P'  =  mv  -{-  triv' 

=  (m  +  m')v  —  \m{lco  -f-  /'co').        (1) 
Taking  moments  about  B  for  the  body  AB,  thus  eliminating  the 
hinge-impulse, 

P'a  =  ml2co/ 12  -f-  tnvl/2.        .         .         .     (2) 

Taking  moments  about  B  for  the  body  BC, 

o  s=  m'l''2(o'/ 12  —  m'v'1' I2  =  (3/ft)  -f-  4.1'co'  —  6v)m'l' 1 12.      (3) 

Equations  (1),  (2)  and  (3)  serve  to  determine  v,  co  and  co'  when 

P'  is  known.     Without  knowing  P\  the  ratios  of  ft),  ft)'  and  v  can 

be  determined. 

2.  If  the  two  bars  are  equal  in  all  respects  and  the  impulse  is 
applied  at  the  middle  point  of  AB,  show  that  co  =  co'  =  6v/jl. 

3.  At  what  point  must  the  impulse  be  applied  in  order  that  the 
bar  BC  shall  be  instantaneously  at  rest?     Will  it  remain  at  rest? 

4.  In  Ex.  1,  suppose  the  two  bars  equal  in  all  respects  and  the 
impulse  applied  at  A.     Show  that  co  —  —  0/3  =  —  6^/5/. 

5.  Two  bodies  A,  B  of  masses  m,  m! ',  are  connected  by  a  string 
which  coincides  with  the  line  joining  their  mass-centers.  At  a  certain 
instant  they  are  moving  with  equal  velocities  at  right  angles  to  the 
string,  when  A  receives  an  impulse  in  the  direction  BA  which,  MA 
were  free,  would  have  deflected  it  450.  Determine  the  motion  just 
after  the  impulse,  assuming  that  the  string  does  not  stretch  nor  break. 

6.  The  initial  conditions  being  as  in  Ex.  5,  let  A  receive  an  im- 
pulse directed  at  an  angle  of  300  from  BA  produced,  of  such  magnitude 
that  if  A  were  free  is  would  be  deflected  150.  Determine  the  motion 
just  after  the  impulse. 

Ans.  If  v  is  the  original  velocity,  the  bodies  have  equal  velocities 
\o.^i6()mj(m  -\-  m'y\v  in  the  direction  BA  after  the  impulse;  the 
velocity  of  B  perpendicular  to  BA  is  unchanged,  while  the  final  ve- 
locity of  A  perpendicular  to  BA  is  1.183Z/. 


CHAPTER   XXIII. 

THEORY    OF    ENERGY. 

§  i.  External  and  Internal  Work. 

482.  Work  of  a  Stress. —  The  forces  which  any  two  particles 
exert  upon  each  other  are  equal  and  opposite,  constituting  a  stress 
(Art.  36).  The  total  work  done  by  the  two  forces  of  a  stress  during 
any  displacements  of  the  particles  upon  which  they  act  may  be  com- 
puted as  follows  : 

Let  A  and  B  (Fig.  190)  be  the  two  particles,  r  their  distance 
apart  at  any  instant,  P  the  magnitude  of 
the  force  which  each  exerts  upon  the 
other.  Assume  that  the  force  acting 
upon  A  has  the  direction  BA  ;  then  the 
force  acting  upon  B  has  the  opposite 
direction  AB. 

Let  the  particles  receive  any  innni-  Fig.  190. 

tesimal  displacements  AA'}   BB',  such 

that  the  angle  turned  through  by  AB  is  also  infinitesimal,  and  let 
A  A" \  BB"  be  the  orthographic  projections  of  these  displacements 
upon  AB.     Then 

—  PX  A  A'  =  work  done  by  force  P  acting  on  A  ; 

PX  BB"=      "       "      "      "     P      "      "  B; 

P{_BB"  —  A  A")  =  total  work  done  by  stress. 

But       BB"  —  AA"  =  A"B"  —  AB  =  AB'  cos  (dd)  —  AB} 

if  dO  is  the  angle  between  AB'  and  AB.     And  since 

cos  (dd)  =  1  —  \{dOy  +     .     .     .     , 

AB'  cos  (dd)  differs  from  AB'  by  an  infinitesimal  of  the  second 
order,  and 

AB'  cos  (dO)  —  AB  =  AB'  —  AB  =  dr. 

The  total  work  done  by  the  stress  during  the  supposed  infinitesimal 
displacement  is  therefore  Pdr ;  and  during  any  finite  displacement 
the  work  is  equal  to 


406  THEORETICAL    MECHANICS. 


I 


r" 

Pdr,  .     (i) 


r*  and  r"  being  the  initial  and  final  values  of  r. 

The  same  result  may  be  reached  by  another  method,  as  follows : 
Let  (jr, ,  yx ,  zx)  and  (x.2 ,  y%,  z2)  be  the  rectangular  coordinates  of  the 
particles  A  and  B  respectively,  and  let  a,  /3,  7  be  the  angles  between 
AB  and  the  axes.  By  Art.  352,  the  work  done  by  either  of  the 
forces  P  is  equal  to  the  sum  of  the  works  done  by  its  axial  com- 
ponents. Hence  the  force  P  acting  upon  A  does  an  amount  of 
work 

— fP  cos  a  •  dxx  — f  P  cos  /3  •  dyx  — fP  cos  7    dz% , 

and  the  force  P  acting  upon  B  does  an  amount  of  work 

f  P  cos  a  -  dx.t  -f-  f  P  cos  $  •  dy.,  +  fP  cos  7  •  dz2 ; 

the  limits  of  the  integrations  being  so  taken  as  to  correspond  to  the 
initial  and  final  positions  of  the  particles.  The  total  work  done  by 
both  the  forces  of  the  stress  is  therefore 

fP[(dx2  —  dxx)  cos  a  -\-  (dy2  —  dyx)  cos  /3  -f-  (dz2  —  dzx)  cos  7]. 
From  the  equation 

r>  =  (x2-xty  +  {,,  -y,f  +  (*,  -  z,f, 
there  results,  by  differentiation, 

rdr  =  Car,  —  xx)(dx2  —  dxx)  +  (7,  —y^{dyx  —  ^) 

-f  fe  —  ^i)(^2  —  dzd ; 

from  which,  since 

(x2  —  *i)/r  =  cos  a,     ( jj/2  —  jO/r  =  cos  A     (^  —  sx)jr  =  cos  7, 

we  have 

dr  =  {dxt  —  dxx)  cos  a  -\-  (dy.2  —  dyx)  cos  ft  -\~  (dz%  —  dzx)  cos  7. 

The  above  value  of  the  total  work  done  by  the  two  forces  of  the  stress 
therefore  reduces  to 

Pdr, 


r 


as  before. 

Two  particles  rigidly  connected. —  If  the  two  particles  A  and  B 
are  rigidly  connected,  or  if  they  move  in  such  a  way  that  the  distance 
between  them  remains  constant,  the  work  done  by  the  stress  is  zero ; 
for  if  r  is  constant,  the  definite  integral  (1)  is  zero,  whatever  the  value 
of  P,  so  long  as  it  is  finite. 


THEORY    OF    ENERGY.  407 

The  above  reasoning  obviously  holds  for  any  two  equal  and  oppo- 
site forces  applied  to  two  particles  and  directed  along  the  line  joining 
them,  whether  the  two  forces  constitute  a  stress,  or  whether  they  are 
exerted  by  other  bodies  or  particles. 

483.  External  and  Internal  Work  Defined. — Work  done  upon 
a  system  of  particles  is  external  if  done  by  an  external  force  (a  force 
exerted  by  a  particle  not  belonging  to  the  system)  ;  it  is  internal  if 
done  by  an  internal  force. 

484.  Configuration. —  A  definite  set  of  relative  positions  of  the 
particles  of  a  system  is  called  a  configuration.  If  all  the  distances 
between  the  particles  remain  constant,  the  configuration  remains 
constant.  If  the  configuration  does  not  change,  the  system  either 
remains  at  rest  or  moves  as  if  the  particles  were  rigidly  connected. 

485.  Internal  Work  Done  Only  During  Change  of  Configura- 
tion.—  Every  internal  force  is  one  of  the  forces  of  an  internal  stress. 
The  work  done  by  a  stress  is  zero  unless  the  particles  between  which 
the  stress  acts  change  their  distance  apart  (Art.  482).  The  total 
work  done  by  the  internal  forces  upon  the  particles  of  a  system  is 
therefore  zero  unless  the  configuration  changes. 

§  2.  Energy  of  Any  System  of  Particles. 

486.  Energy  of  a  System  Defined.*  —  When  the  condition  of 
a  system  is  such  that  it  can  do  work  against  external  forces,  it  is  said 
to  possess  energy. 

The  quantity  of  energy  of  a  system  is  the  quantity  of  external 
work  it  will  do  in  passing  from  its  present  condition  to  some  standard 
condition. 

The  meaning  of  ' '  condition ' '  will  be  made  clear  by  the  following 
discussion  of  the  method  of  estimating  the  quantity  of  energy. 

487.  Kinetic  Energy  of  a  System  of  Particles. —  It  has  been 
shown  (Art.  358)  that  a  particle  of  mass  m  and  velocity  v  possesses 
energy  of  motion  or  kinetic  energy  to  the  amount  imv2,  the  "stand- 
ard  condition ' '  being  one  of  rest.     A  system  of  particles  of  masses 

*  The  meaning  of  energy  of  a  system  has  already  been  explained  in  an 
elementary  manner  in  Chapter  XVII  (Arts.  360-366).  Some  of  the  defini- 
tions and  principles  there  given  are  here  repeated  in  order  to  make  the 
following  more  rigorous  discussion  complete. 


4<d8  theoretical  mechanics. 

miy  m2i     .     .     .     ,  having  velocities  vl}  v2,     .     .     .     ,  possesses 
an  amount  of  kinetic  energy 

\m{0?  -f  \n&}  +     .     .     .     . 

For  if  the  particles  are  brought  to  rest  by  the  action  of  external 
forces,  this  amount  of  external  work  will  be  done. 

488.  Energy  of  Configuration,  or  Potential  Energy. —  If  any 
particle  of  a  system  is  acted  upon  by  both  external  and  internal 
forces,  it  may  do  external  work  without  any  change  of  velocity.  For 
its  velocity  will  remain  unchanged  so  long  as  the  internal  work  done 
upon  it  is  equal  to  the  external  work  done  by  it. 

It  thus  appears  that  a  system  may  be  able  to  do  external  work 
(and  therefore  may  possess  energy)  by  reason  of  the  internal  forces. 
Since  the  total  internal  work  is  zero  for  any  displacement  which  leaves 
the  configuration  unchanged,  this  form  of  energy  depends  upon  the 
possibility  of  changing  the  configuration.  It  may  therefore  be  called 
energy  of  configuration.  The  name  potential  energy  is,  however, 
more  generally  used. 

489.  Meaning  of  Standard  Condition. —  The  meaning  of  the 
word  "condition"  used  in  the  definition  of  energy  now  becomes 
clear.  By  a  definite  condition  of  a  system  is  meant  a  definite  con- 
figuration together  with  a  definite  set  of  velocities. 

In  order  to  estimate  the  total  amount  of  energy  possessed  by  a 
system  in  a  given  condition,  it  is  necessary  to  assume  a  standard 
configuration  and  a  standard  set  of  velocities ;  and  to  compute  the 
total  quantity  of  external  work  done  by  the  particles  while  the  sys- 
tem passes  from  the  given  condition  to  this  standard  condition. 

In  estimating  the  kinetic  energy  of  a  particle,  any  velocity  may 
be  taken  as  the  standard,  but  it  will  here  be  assumed  zero  for  every 
particle.  This  simplifies  the  algebraic  expression  for  kinetic  energy, 
and  does  not  detract  from  the  generality  of  the  discussion. 

The  choice  of  the  standard  configuration  will  be  governed  by 
convenience  in  each  particular  case. 

490.  Total  Energy  of  a  System. — The  foregoing  principles  will 
now  be  expressed  in  definite  mathematical  form. 

Let  it  be  required  to  compute  the  total  external  work  done  by  a 
system  in  passing  from  any  given  condition  to  an  assumed  standard 
condition. 


THEORY    OF    ENERGY.  409 

Let  v\  =  velocity  of  particle  of  mass  mx , 

Jx  =  work  done  upon  the  particle  by  internal  forces, 
El  =  work  done  by  the  particle  against  external  forces. 

If  the  particle  is  brought  to  rest, 

Et  —  I1  =  \mxv?. 

A  similar  equation  may  be  written  for  every  particle  ;  that  is, 

E2  —  I2  =  \m%v* ; 


By  addition, 

(fig+mEt+  .   .  .  y- &+;!*+.  .  .  .  ) 

=  i  ^hvx  +  4«W  + 

Or,  if  E  =  Ex  +  E2  +     .     .  .     , 

/  =  /x  +  /,  +     .     .  .     , 

K  =  \m{0?  -\r  \m2v2   +     .     .     .     , 

the  equation  may  be  written 

E  —  I+K. 

Since  E  denotes  the  total  external  work  done  by  all  the  particles 
while  the  system  passes  to  the  standard  condition,  it  is  by  definition 
(Art.  486)  the  total  energy  of  the  system  in  its  initial  condition.  Of 
the  two  parts  I  and  K  whose  sum  is  the  total  energy,  the  second  is 
the  kinetic  energy  of  the  system,  being  equal  to  the  sum  of  the 
kinetic  energies  of  the  individual  particles.  The  part  I  is  the  total 
work  done  by  the  internal  forces  while  the  system  passes  to  the 
standard  condition.  This  is  the  energy  of  configuration  or  potential 
energy.  In  regard  to  this  quantity  an  important  question  must  be 
raised. 

491.  Is  Energy'a  Definite  Quantity? — A  system  may,  in  gen- 
eral, change  from  its  present  condition  to  the  standard  condition  in 
different  ways.  Can  it  be  assumed  that  the  quantity  of  external  work 
done  by  the  system  will  be  the  same,  whichever  one  of  the  possible 
ways  is  actually  followed  ?  If  different  ways  of  making  the  change 
result  in  different  quantities  of  external  work,  the  energy  of  the  sys- 
tem, as  the  term  has  been  defined,  cannot  have  a  definite  value. 

The  kinetic  energy  obviously  has  a  definite  value,  depending  only 
upon  the  velocities  of  the  particles  and  their  masses.  It  remains  to 
consider  the  potential  (or  configuration)  energy.     And  since  the  po- 


4IO  THEORETICAL    MECHANICS. 

tential  energy  is  equal  to  the  total  work  done  by  the  internal  forces 
during  the  change  to  the  standard  configuration  (Art.  490),  the  ques- 
tion reduces  to  this:  Does  the  total  work  done  by  the  internal  forces 
depend  only  upon  the  initial  and  final  configiirations,  or  does  it  de- 
pend upon  the  way  in  which  the  change  of  configuration  takes  place? 

In  the  case  of  any  actual  material  system,  this  question  can  be 
answered  only  by  experience.  The  foregoing  discussion  has  referred 
to  ideal  systems  consisting  of  material  particles  exerting  upon  one 
another  forces  which,  for  any  two  particles,  act  along  the  line  joining 
the  particles,  and  follow  Newton's  law  of  action  and  reaction.  No 
assumption  has  been  made  as  to  whether  the  magnitude  of  the  stress 
acting  between  any  two  particles  depends  upon  their  distance  apart, 
or  upon  their  relative  velocities,  or  upon  other  conditions. 

It  is  possible,  without  contradicting  any  dynamical  law  heretofore 
assumed,  to  assign  laws  of  force  which  shall  make  the  internal  work 
depend  only  upon  the  total  change  of  configuration ;  it  is  equally 
possible  to  assign  laws  which  shall  make  the  internal  work  depend 
upon  the  way  in  which  the  change  of  configuration  takes  place. 

492.  Conservative  and  Non-conservative  Systems.—  Ideally, 
then,  two  kinds  of  material  systems  may  exist. 

If  the  total  internal  work  done  during  any  change  of  configuration 
depends  only  upon  the  initial  and  final  configurations,  the  system  is 
called  conservative. 

If  the  total  internal  work  depends  upon  the  way  in  which  the 
change  of  configuration  is  made,  the  system  is  non- conservative. 

It  is  only  for  conservative  systems,  as  thus  defined,  that  energy  is 
a  definite  quantity. 

493.  Law  of  Force  in  Conservative  System. —  It  may  be  shown 
that  if  the  stress  acting  between  «ny  two  partides  of  the  system  is 
directed  along  the  line  joining  the  particles,  and  if  its  magnitude 
depends  only  upon  their  distance  apart,  the  system  is  conservative. 

It  was  shown  in  Art.  482  that  the  work  done  by  the  stress  be- 
tween any  two  particles  is  equal  to 


/: 


Pdr, 


if  P  is  the  magnitude  of  each  of  the  equal  and  opposite  forces  of 
the  stress,  r  the  distance  between  the  particles,  r'and  r"the  initial 


THEORY    OF    ENERGY.  4II 

and  final  values  of  r.  If  P  is  a  function  of  r,  f  Pdr  is  a  function  of  r, 
and  the  definite  integral  is  a  function  of  r '  and  r" .  Since  the  internal 
forces  consist  wholly  of  stresses,  the  total  internal  work  is  a  function 
of  the  initial  and  final  values  of  the  distances  between  the  particles ; 
that  is,  it  depends  only  upon  the  initial  and  final  configurations.  The 
assumed  law  of  force  therefore  makes  the  system  conservative. 

The  above  assumption  is  not,  however,  the  only  one  which  makes 
the  system  conservative.  Considering  all  possible  pairs  of  particles, 
let  Pl  be  the  magnitude  of  the  stress  acting  between  a  pair  whose 
distance  apart  is  rx ;  P2  the  magnitude  of  the  stress  between  a  pair 
whose  distance  apart  is  r2 ;  etc. ;  and  let  /  denote  the  total  work 
done  by  the  internal  forces  during  any  given  change  of  configuration. 
Then  (Art.  482) 

dl =  Pxdrx  +  P2dr2  +     .... 

If  Plt  P2,  .  .  .  are  such  functions  of  rx ,  r2  .  .  .  that  the 
second  member  of  this  equation  is  the  differential  of  a  function  of 
rltrt,  .  .  .  ,  the  integration  of  the  equation  between  proper 
limits  gives  /as  a  function  of  the  initial  and  final  values  of  rls  r2, 
.  .  .  ,  that  is,  as  a  function  of  the  initial  and  final  configurations. 
As  an  illustration,  a  system  of  three  particles  is  conservative  if 
the  law  of  force  is  expressed  by  the  equations 

Pl==A  +  £(rt  +  rs)  +  Cr2r.s; 
P2  =  A  -{-  B(r,  +  rj  +  Crtrt; 
P3  =  A  +B(rx  +  r±+  Crtr9l 

A,  B  and  C  being  constants. 

Examples. 

1 .  Two  particles  attract  each  other  with  a  stress  of  constant  mag- 
nitude equal  to  100  poundals.  What  is  the  potential  energy  of  the 
system  when  the  particles  are  10  ft.  apart,  if  in  the  standard  config- 
uration they  are  4  ft.  apart  ? 

2.  Two  particles  attract  each  other  with  forces  varying  inversely 
as  the  square  of  their  distance  apart,  the  value  of  the  attraction  being 
c  when  the  distance  is  a.  If  the  distance  is  b  in  the  standard  config- 
uration, what  is  the  potential  energy  of  the  system  when  the  distance 
is  r?  Ans.  ca\i/d  —  i/r). 

3.  A  system  consists  of  three  particles  attracting  one  another 
according  to  the  law  of  the  inverse  square  of  the  distance.  The 
attraction  between  any  two  when  1  met.  apart  is  5,000  dynes.     In 


412  THEORETICAL    MECHANICS. 

the  standard  configuration  their  distances  are  2,  3  and  4  met.  re- 
spectively. Determine  the  value  of  the  potential  energy  when  each 
of  the  three  distances  is  10  met.  Ans.  3.92  X  io5  ergs. 

494.  Principle  of  Work  and  Energy  for  Conservative  System. — 
Let  a  conservative  system  experience  any  change  of  condition,  not 
necessarily  passing  to  the  standard  condition.  The  initial  and  final 
conditions  may  be  designated  briefly  as  Cx  and  C2 ,  and  the  standard 
condition  as  C0.  In  passing  from  Cx  to  C0  in  any  way  the  total 
external  work  is  a  definite  quantity  Ex ,  equal  by  definition  to  the 
energy  of  the  system  in  the  condition  Cx .  In  passing  from  C2  to 
C0  in  any  way  the  total  external  work  is  a  definite  quantity  E2 ,  equal 
to  the  energy  of  the  system  in  the  condition  C2.  Hence  in  passing 
from  Cx  to  C2  ,  the  external  work  done  by  the  system  is  a  definite 
quantity  E1  —  E2 .     That  is, 

The  total  external  work  done  by  a  conservative  system  during 
any  change  of  condition  is  equal  to  the  decrease  of  its  total  energy. 

The  same  principle  may  be  stated  in  the  following  form  : 

The  total  work  done  upon  a  conservative  system  by  extei'na  I forces 
during  any  change  of  condition  is  equal  to  the  increase  of  its  total 
energy. 

As  a  particular  case,  suppose  a  change  in  which  no  external  work 
is  done.  The  total  energy  must  remain  constant,  the  potential  and 
kinetic  energies  receiving  opposite  and  equal  changes. 

In  the  following  Section  will  be  given  a  brief  discussion  of  the 
theory  of  energy  as  applied  to  the  actual  systems  of  nature. 

§  3.    Conservation  of  Energy. 

495.  Transfer   of  Energy  from   One   System   to   Another. — 

When  a  conservative  system  does  work  against  external  forces,  it 
loses  an  amount  of  energy  equal  to  the  work  done.      In  certain  cases 
this  loss  is  accompanied  by  a  gain  of  energy  on 
the  part  of  some  other  body  or  system. 

Thus,  consider  two  bodies  A  and  B  (Fig.  191) 

to  be  in  contact  and  to  be  exerting  upon  each 

other  equal  and  opposite  forces  at  the  surface  of 

Fig.  191.  contact,  the  magnitude  of  each  force  being  P,  and 

its  direction  being  normal  to  the  surface  of  contact. 

Let  the  bodies  receive  displacements  whose  components  parallel  to 


B 

A 

Py 

C/> 

THEORY    OF    ENERGY.  413 

the  forces  are  equal,  each  being  equal  to  h,  and  their  direction  being 
such  that  A  does  positive  work  and  B  therefore  does  negative  work. 
The  body  A  loses  energy  to  the  amount  Ph,  and  the  body  B  gains 
an  equal  amount.  The  action  between  the  two  bodies  has  therefore 
resulted  in  no  change  in  their  total  amount  of  energy ;  one  has 
gained  exactly  as  much  as  the  other  has  lost.* 

Again,  consider  two  rough  bodies  A  and  B  in  contact  (Fig.  192), 
exerting  upon  each  other  forces  which  have  components  both  normal 
and  tangential  to  the  surface  of  contact.  The  normal  components 
are  equal  and  opposite  forces  JV,  and  the  tangential  components  are 
equal  and  opposite  forces  T.  Let  both  bodies  be  displaced  parallel 
to  the  forces  T ;  let  h '  be  the  displace- 
ment of  A  and  h  "  the  displacement  of 
B,  their  directions  being  the  same,  and  — ^ 
h'  being  greater  than  h" .     No  work  is 


\tA 


N-     h' 


T      7\       \A  .T 


NB 


"> 


N 


N 


^ 


done  by  or  against  either  of  the  normal  I h' 

forces.     The  forces  T  act  in  such  di-  Fig.  192. 

rections  as  to  resist  the  sliding  of  one 

body  over  the  other  ;  that  is,  the  direction  of  the  force  T  acting 
upon  A  is  opposite  to  the  direction  of  the  displacement  of  its  point 
of  application,  while  the  force  T  acting  upon  B  acts  in  the  direction 
of  the  displacement  of  its  point  of  application. 

The  body  A  does  work  equal  to  -f-  Th ',  and  loses  energy  equal 
to  +  Th'. 

The  body  B  does  work  equal  to  —  Th" ,  and  gains  energy  equal 
to  -f  Th". 

The  body  A  therefore  loses  more  energy  than  B  gains.  The 
quantity  of  energy  T{Ji   —  h")  has  apparently  disappeared. 

If,  in  the  last  case,  h'  =  h",  so  that  no  sliding  occurs,  the  energy 
gained  by  B  is  equal  to  that  lost  by  A.  If  h"  =  o,  B  gains  no  en- 
ergy, and  there  is  an  apparent  loss  of  all  the  energy  given  up  by  A. 

Another  example  of  an  apparent  loss  of  energy  is  furnished  by  a 
moving  body  suddenly  brought  to  rest,  as  a  falling  body  striking  the 
earth.  The  kinetic  energy  possessed  by  the  body  just  before  strik- 
ing is  wholly  lost,  without  any  apparent  gain  by  other  bodies. 


*  It  should  be  noticed  that  nothing  is  here  said  of  the  relation  of  A  or  of 
B  to  other  bodies.  The  energy  received  by  B  from  A  may  be  continually 
transferred,  in  whole  or  in  part,  to  other  bodies. 


414  THEORETICAL    MECHANICS. 

An  example  of  the  transfer  of  energy  from  one  system  to  another 
is  furnished  by  a  body  moving  upward  against  the  force  of  gravity. 
A  body  of  mass  m,  moving  upward  with  velocity  v,  possesses  kinetic 
energy  equal  to  \mv2.  As  it  rises,  its  kinetic  energy  decreases  and 
finally  is  wholly  lost.  Is  this  loss  accompanied  by  a  gain  on  the 
part  of  other  bodies  or  systems  ?  The  other  body  concerned  is  the 
earth  ;  obviously  the  system  consisting  of  the  earth  and  the  body 
has  gained  an  amount  of  potential  energy  equal  to  mgh,  if  h  is  the 
vertical  distance  ascended  by  the  body.  If,  during  this  ascent,  the 
velocity  of  the  body  changes  from  vx  to  v2 ,  its  loss  of  kinetic  energy 
is  \m(y?  —  vf).  But  mgh  and  im(v*  —  v2l)  are  equal  (Art.  227); 
that  is,  the  gain  of  potential  energy  by  the  system  of  the  earth  and 
the  body  is  exactly  equal  to  the  loss  of  kinetic  energy  by  the  body. 

The  last  example  is  an  illustration  of  what  occurs  whenever  a 
conservative  system  suffers  a  change  of  configuration  without  doing 
external  work.  The  individual  members  of  the  system  may  gain  or 
lose  kinetic  energy,  but  the  system  at  the  same  time  loses  or  gains 
an  equal  amount  of  potential  energy.  If,  in  the  supposed  case,  the 
earth  and  the  body  be  regarded  as  a  system,  their  energy  does  not 
vary  in  total  amount,  but  merely  changes,  in  part  from  kinetic  to 
potential. 

To  summarize:  When  a  body  or  a  system  loses  energy,  there 
may  be  (a)  an  equal  gain  by  other  bodies  or  systems ;  (b)  a  gain  by 
other  systems  apparently  less  than  the  loss  by  the  given  system  ;  or 
(c)  no  apparent  gain  by  other  systems. 

In  some  cases,  then,  there  seems  to  be  a  disappearance  of  energy. 

496.  Equivalence  of  Energy  and  Heat. —  In  many  cases  in 
which  energy  disappears,  heat  is  generated,  and  it  is  found  that  there 
is  a  definite  relation  between  the  quantity  of  energy  that  disappears 
and  the  quantity  of  heat  that  is  generated.  Taking  as  the  unit  quan- 
tity of  heat  the  amount  which  will  raise  the  temperature  of  a  pound 
of  water  one  degree  Fahr.  (the  British  thermal  unit),  the  disappear- 
ance of  778  foot-pounds  of  energy  is  accompanied  by  the  generation 
of  1  unit  of  heat.  * 

*To  make  the  unit  heat  definite,  the  temperature  of  the  water  should  be 
specified,  since  the  quantity  of  heat  required  to  raise  the  temperature  of  a 
pound  of  water  one  degree  is  not  exactly  the  same  at  all  temperatures.  The 
numerical  value  of  the  energy-equivalent  of  the  unit  heat,  expressed  in  foot- 


THEORY   OF    ENERGY.  415 

In  other  cases,  as  in  the  steam-engine,  energy  comes  into  exist- 
ence while  heat  disappears  ;  one  unit  of  heat  being  consumed  for 
every  778  foot-pounds  of  energy  generated. 

From  these  facts  it  is  natural  to  conclude  that  heat  and  energy 
are  equivalent,  and  that  a  given  quantity  of  one  may  be  transformed 
into  an  equivalent  quantity  of  the  other. 

497.  Forms  of  Energy. —  Modern  science  goes  further  and  re- 
gards heat  not  merely  as  equivalent  to  energy  but  as  being  actually  a 
form  of  energy.  Moreover,  heat  is  but  one  of  several  forms  of  en- 
ergy into  which  the  ordinary  kinetic  and  potential  energy  possessed 
by  bodies  and  systems  may  be  converted,  either  directly  or  indirectly. 
The  meaning  of  the  word  energy  is  thus  enlarged ;  and  we  recog- 
nize mechanical  energy,  heat  energy,  chemical  energy,  electrical 
energy. 

Mechanical  energy. —  Energy  which  depends  upon  the  config- 
uration of  a  material  system  or  upon  the  motions  of  its  constituent 
particles  is  mechanical  energy.  The  discussion  included  in  Arts. 
486-494  refers  solely  to  this  kind  of  energy. 

Heat  energy.  —  The  reasons  for  regarding  heat  as  a  form  of  en- 
ergy are  given  above. 

Chemical  energy. —  The  chemical  combination  of  two  bodies  is 
often  accompanied  by  the  generation  of  heat.  In  this  case  there  is 
found  to  be  a  definite  relation  between  the  amount  of  chemical  change 
and  the  quantity  of  heat  generated.  Thus,  if  one  pound  of  water  is 
produced  by  the  combination  of  hydrogen  and  oxygen,  the  quantity 
of  heat  generated  is  6,900  British  thermal  units.  Again,  by  the  ex- 
penditure of  this  amount  of  heat  (or  of  its  equivalent  in  other  forms 
of  energy),  a  pound  of  water  may  be  decomposed  into  hydrogen  and 
oxygen.  In  such  a  case  as  this  the  two  bodies  when  uncombined 
may  be  said  to  possess  chemical  energy. 

Electrical  energy.  —  Another  important  form  of  energy  is  pos- 
sessed by  bodies  by  reason  of  their  electrical  condition.     If  a  body 

pounds,  varies  with  the  value  of  the  pound-force,  which  varies  with  the  local- 
ity. The  value  778  foot-pounds  is  sufficiently  correct  for  most  purposes. 
Expressed  in  British  absolute  units  of  work  (foot-poundals)  the  energy- 
equivalent  of  the  unit  heat  is  approximately  25,000. 

Using  French  units,  the  quantity  of  heat  required  to  raise  the  tempera- 
ture of  a  kilogram  of  water  a  degree  Cent,  is  equivalent  to  about  427  meter- 
kilograms  of  energy. 


41 6  THEORETICAL    MECHANICS. 

is  charged  with  electricity,  and  if  the  conditions  are  such  that  there 
can  be  a  flow  of  electricity  accompanied  by  a  fall  of  potential,* 
the  body  possesses  electrical  energy.  If  the  flow  actually  takes 
place,  the  electrical  energy  is  transformed  into  energy  of  some  other 
form.  In  the  case  of  an  electric  current,  there  is  a  continuous  flow 
of  electricity  and  a  fall  of  potential  in  the  direction  of  the  flow.  A 
part  of  the  electrical  energy  is  transformed  into  heat  which  manifests 
itself  in  the  heating  of  the  conductor  carrying  the  current.  A  part 
may  be  transformed  into  mechanical  energy  by  means  of  a  motor. 
Or  the  current  may  be  employed  in  decomposing  a  chemically  com- 
pound substance  as  water  into  its  elements,  a  part  of  the  electrical 
energy  being  then  transformed  into  chemical  energy. 

Nature  of  different  forms  of  energy.  —  The  nature  of  each  of 
the  above  forms  of  energy  is  unknown ;  any  hypothesis  concerning 
it  must  involve  a  theory  as  to  the  ultimate  constitution  of  matter.  A 
plausible  supposition  is  that  heat  and  chemical  energy  are  dependent 
upon  the  motions  and  relative  positions  of  the  ultimate  particles  of 
which  bodies  are  made  up.  On  this  supposition,  these  forms  of  en- 
ergy do  not  differ  in  nature  from  the  kinetic  and  potential  energy 
possessed  by  any  system  by  virtue  of  its  configuration  and  of  the 
visible  motions  of  its  parts.  Thus,  the  accepted  theory  regarding 
the  nature  of  heat  is  that  it  is  kinetic  energy  possessed  by  the  mo- 
lecules of  bodies ;  and  a  possible  explanation  of  the  development  of 
heat  during  the  chemical  combination  of  two  substances  is  that  there 
is  a  transformation  of  mechanical  energy  from  potential  into  kinetic. 
The  atoms  of  hydrogen  and  those  of  oxygen,  when  brought  into 
certain  relations,  may  exert  upon  each  other  forces,  under  the  action 
of  which  the  particles  approach  one  another  or  assume  definite  rela- 
tive positions ;  during  this  process  work  may  be  done  upon  the 
particles  by  these  forces,  and  their  kinetic  energy  may  thus  be  in- 
creased, the  molecular  kinetic  energy  being  heat.  Before  the  com- 
bination the  two  bodies,  regarded  as  a  system,  possess  energy  by 
virtue  of  the  relative  positions  of  their  particles  and  of  the  forces 
exerted  between  them.  If  this  is  the  true  explanation,  the  energy, 
both  before  and  after  the  combination  takes  place,  is  of  the  same 


*  No  explanation  of  the  meaning  of  electrical  potential  can  be  given  here. 
The  above  brief  statement  regarding  electrical  energy  presupposes  an  ele- 
mentary knowledge  of  the  theory  of  electricity. 


THEORY    OF    ENERGY.  4J7 

nature  as  ordinary  mechanical  energy ;  being  at  first  potential,  and 
then  transformed  into  kinetic. 

Attempts  have  been  made  to  explain  electrical  energy  as  me- 
chanical energy  possessed  by  the  hypothetical  "ether." 

The  supposition  that  all  energy  is  really  mechanical  may  or  may 
not  become  more  probable  with  the  advance  of  knowledge  ;  the  fact 
that  the  different  forms  of  energy  are  equivalent  in  the  sense  that  a 
definite  quantity  of  one  can  be  converted  into  a  definite  quantity  of 
another  is  independent  of  any  hypothesis  as  to  their  nature. 

498.  Conservation  of  Energy. —  One  of  the  most  important  re- 
sults of  modern  science  is  the  establishment  of  the  principle  that  no 
system  of  bodies  gains  or  loses  energy  wihout  an  equivalent  loss  or 
gain  by  other  bodies  or  systems. 

499.  Conservation  of  Energy  in  Machines. —  A  machine  is  a 
structure  designed  to  do  work  against  external  forces. 

In  doing  external  work,  a  machine  loses  an  amount  of  energy 
equal  to  the  work  done.  Energy  must  therefore  be  supplied  to  it  if 
it  is  to  continue  to  do  work  indefinitely.  The  operation  of  a  machine 
thus  involves  (a)  the  transfer  of  energy  to  the  machine  from  other 
bodies,  and  (Jj)  the  transfer  of  energy  from  the  machine  to  other 
bodies.  In  general,  these  two  processes  go  on  simultaneously  at 
nearly  the  same  rate.  If  they  go  on  at  unequal  rates,  the  quantity 
of  energy  possessed  by  the  machine  fluctuates. 

If  energy  is  received  by  the  machine  and  given  off  by  it  at  the 
same  rate,  the  machine  is  a  body  or  system  in  equilibrium.  In  this 
case  the  relations  which  must  hold  among  the  forces  acting  upon  it 
may  be  determined  by  the  principles  of  Statics.  Machines  have  been 
considered  from  this  point  of  view  in  Part  I.  The  theory  of  energy 
furnishes  a  means  of  dealing  with  machines  in  a  more  general  manner. 

In  the  light  of  the  principle  of  the  conservation  of  energy,  a 
machine  may  be  defined  as  a  structure  designed  to  transfer  energy 
from  one  body  or  system  to  another. 

500.  Utilized  Energy  and  Lost  Energy. —  That  part  of  the  en- 
ergy given  up  by  a  machine  which  is  equivalent  to  the  useful  work 
done  may  be  called  utilized  energy.  The  energy  given  up  by  the 
machine  can  never  be  wholly  utilized.  In  order  that  the  parts  of  the 
machine  may  move  in  the  desired  manner,  they  must  be  guided  by 
other  bodies ;  these  exert  frictional  forces,  against  which  work  must 


41 8  THEORETICAL   MECHANICS. 

be  done.  In  doing  work  against  the  frictional  forces,  energy  is  trans- 
formed into  heat.  Such  energy  may  be  called  lost,  since  no  useful 
effect  can  be  obtained  from  it. 

The  external  forces  acting  upon  a  machine  may  be  classified  (as 
in  Art.  in)  as  efforts,  resistances  (useful  and  prejudicial),  and 
constraints.  In  terms  of  work  and  energy  these  may  be  defined  as 
follows : 

An  effort  is  a  force  which  does  positive  work  upon  the  machine. 
The  machine  receives  a  quantity  of  energy  equal  to  the  work  done. 

A  resistance  is  a  force  which  does  negative  work  upon  the  ma- 
chine (or  against  which  the  machine  does  positive  work).  The 
machine  gives  up  a  quantity  of  energy  equal  to  the  work  done. 
Energy  given  up  in  doing  work  against  useful  resistances  is  utilized ; 
it  is  lost  if  given  up  in  doing  work  against  prejudicial  resistances. 

A  constraint  is  a  force  which  does  no  work,  positive  or  negative, 
upon  the  machine.  The  office  of  a  constraint  is  to  guide  the  motion 
of  some  machine  part.  No  energy  is  gained  or  lost  by  the  machine 
by  reason  of  a  constraint. 

501.  Equation  of  Energy  for  a  Machine. —  The  transformations 
of  energy  with  which  a  machine  is  concerned  during  a  given  time 
may  be  expressed  algebraically,  in  the  the  most  general  case,  as 
follows : 

Let  W  =  work  done  upon  the  machine  by  the  efforts 
=  energy  received  by  the  machine ; 
W'  =  work  done  by  the  machine  against  useful  resistances 

—  energy  utilized ; 
W"  =a  work  done  by  the  machine  against  prejudicial  resistances 
=  energy  lost ; 
E  =  increase  in  the  quantity  of  energy  possessed  by  the 
machine. 
Then  W  =  W  +  W"  +  E. 

If  the  energy  possessed  by  the  machine  remains  constant,  the 
equation  is 

W =  W  +  W". 

The  same  equation  holds  if  the  interval  is  so  chosen  that  the  energy 
possessed  by  the  machine  has  the  same  value  at  the  end  as  at  the 
beginning  of  the  interval,  even  though  it  has  fluctuated  during  the 
interval. 


THEORY    OF    ENERGY.  419 

502.  Efficiency. —  For  the  efficient  working  of  a  machine,  the 
energy  lost  is  to  be  made  as  small  as  possible. 

The  efficiency  of  a  machine  is  the  ratio  of  the  energy  utilized  to 
the  energy  received  by  the  machine,  the  energy  stored  in  the  machine 
being  supposed  to  remain  constant  (or  the  interval  of  time  to  be  so 
taken  that  the  total  energy  lost  by  the  machine  has  equaled  the  total 
energy  received). 

Consider  an  interval  such  that  the  energy  stored  in  the  machine 
has  the  same  value  at  the  beginning  and  at  the  end  of  the  interval, 
so  that  E  =  o.     The  efficiency  may  be  expressed  by  the  equation 

e=  W/W=  W/(W  +  W"). 

If  W"  could  be  made  zero,  the  efficiency  would  be  i  ;  its  actual  value 
is  always  less  than  i. 

Examples. 

1.  The  fuel  used  in  running  a  steam-engine  is  coal  of  such  com- 
position that  the  combustion  of  i  lb.  produces  heat  sufficient  to  raise 
the  temperature  of  12,000  lbs.  of  water  i°  Fahr.  It  is  found  that  3^ 
lbs.  of  fuel  are  consumed  per  horse-power  per  hour.  What  is  the 
efficiency  of  the  entire  apparatus  ?  Ans.  0.061  nearly. 

2.  A  steam-engine  uses  coal  of  such  composition  that  the  com- 
bustion of  1  lb.  generates  10,000  British  thermal  units.  If  40  lbs.  of 
coal  are  used  per  hour,  and  if  the  efficiency  is  0.08,  what  horse-power 
is  realized  ? 

3.  A  dynamo  is  driven  by  an  engine  working  as  described  in 
Ex.  2.  If  its  efficiency  is  0.78,  what  "  activity"  in  kilowatts  is  repre- 
sented by  the  current  generated  ? 

§  4.   Principle  of  Work  and  Energy  Applied  to  Rigid  System. 

503.  Potential  Energy  of  Rigid  System. —  It  was  shown  in 
Art.  485  that  internal  work  is  done  only  when  the  configuration  of 
the  system  changes.  In  any  possible  displacement  of  a  rigid  system, 
therefore,  the  internal  work  is  zero. 

It  follows  that  the  potential  energy  of  a  rigid  system  is  zero. 

504.  Kinetic  Energy  of  Rigid  System. —  When  the  condition 
of  motion  of  a  rigid  system  at  any  instant  is  known,  the  value  of  the 
kinetic  energy  can  be  expressed  in  a  simple  manner. 

(a)   Translation. —  In  case  of  translation  the  velocities  of  all  par- 
ticles are  equal  in  magnitude.     The  kinetic  energy  is  therefore 


420  THEORETICAL    MECHANICS. 

K  =  \mxv?  +  \m{o?  +  .  .  .  =  \Mv\  .  (i) 
if  M  is  the  total  mass  of  the  system  and  v  the  velocity  of  every 
particle. 

(J?)  Rotation  about  fixed  axis. —  Let  rlt  r%i  .  .  .  be  the 
distances  from  the  axis  of  rotation  of  particles  whose  masses  are 
mlt  m%y  .  .  .  ;  /  the  moment  of  inertia  of  the  system  with  re- 
spect to  the  axis  of  rotation  ;  (o  the  angular  velocity  at  any  instant. 
The  velocities  of  the  several  particles  have  values  rx(o,  rzco}  .  .  .  , 
and  the  kinetic  energy  is  therefore 

K  =  \mx{rx<x>f  -f-  im£r,<oy  +     .     .     . 

=  Urnf*  +  **«  r2  +    •    •    •  K ; 

or  K=iI<o\  ....     (2) 

(V)  Any  plane  motion. —  Since  any  plane  motion  of  a  rigid  body 
is  at  every  instant  equivalent  to  a  rotation  about  a  definite  axis,  the 
kinetic  energy  in  any  case  (except  translation)  is  given  by  equation 
(2),  if  /  is  the  moment  of  inertia  with  respect  to  the  instantaneous 
axis.  If  the  axis  of  rotation  is  fixed,  /  is  constant ;  but  in  general 
/  is  variable. 

Let  M  =  total  mass  of  system  ; 

a  =  distance  of  mass-center  from  instantaneous  axis ; 

I0  =  moment  of  inertia  with  respect  to  central  axis  parallel 

to  instantaneous  axis  ; 
v  =  aw  =  velocity  of  mass-center. 

From  Art.  400,  /  =  I0  +  Ma1 ; 

therefore  K  =  \M  =  \iy  +  \Mv\       .         .         .     (3) 

The  first  term  of  this  value  of  K  would  be  the  value  of  the  kinetic 
energy  of  the  system  if  rotating,  with  its  actual  angular  velocity, 
about  a  fixed  axis  through  the  mass-center.  The  second  term  would 
be  the  value  of  the  kinetic  energy  if  every  particle  had  a  velocity 
equal  to  that  of  the  mass-center.  These  two  quantities  may  be  called 
rotational  energy  and  translational  energy  respectively. 

505.  Principle  of   Work    and   Energy   for   Rigid   System.— 

Since  the  potential  energy  of  a  rigid  system  is  zero,  the  principle  of 
work  and  energy  takes  the  following  form  : 

The   total  work  done  \     I    \a  rigid  system  is  equal  to  the 

\  decrease  \  of  its  kinetic  energy. 
(  increase  j  J  °y 


THEORY    OF    ENERGY.  42 1 

506.  Equation  of  Work  and  Energy  in  Case  of  Translation.— 

Let  W  denote  the  total  work  done  upon  a  rigid  system  by  all  ex- 
ternal forces  during  a  certain  interval,  and  let  v'  and  v"  be  the  initial 
and  final  values  of  the  velocity  of  the  mass-center.  If  the  motion 
is  a  translation,  the  equation  of  work  and  energy  is 

W  =\M(y'"1  —  O,  •         •         •     (4) 

which  is  identical  with  the  equation  of  work  and  energy  for  a  par- 
ticle of  mass  M  moving  with  the  center  of  mass  of  the  system. 

The  same  equation  applies  to  any  case  in  which  the  angular 
velocity  remains  constant ;  for  equation  (3)  shows  that  if  a>  is  constant 
the  whole  change  in  K  is  equal  to  the  change  in  the  term  ^Mv2. 
Equation  (4)  may  therefore  be  applied  to  any  case  in  which  the  re- 
sultant of  the  external  forces  is  a  single  force  whose  line  of  action 
passes  through  the  mass-center  of  the  body  ;  since  the  angular  veloc- 
ity is  constant  when  this  condition  is  fulfilled  (Art.  444). 

In  the  case  of  translation,  the  work  done  by  any  force  may  be 
computed  as  if  the  force  were  applied  at  the  mass-center,  since  its 
actual  point  of  application  receives,  in  any  interval,  a  displacement 
equal  and  parallel  to  that  of  the  mass-center. 

In  the  case  of  uniform  angular  velocity,  the  actual  displacement 
of  the  point  of  application  of  each  force  must  be  used  *  in  comput- 
ing W. 

507.  Equation  of  Work  and  Energy  in  Case  of  Rotation. — 
If  a  body  rotates  about  a  fixed  axis,  equation  (2)  gives  the  value  of 
the  kinetic  energy  at  any  instant.  If  o>'  and  on"  are  the  initial  and 
final  values  of  the  angular  velocity,  the  equation  of  work  and  energy 
may  be  written  in  the  form 

w=\ii<s*-*y      ...    (5) 

508.  Computation  of  Work  in  Case  of  Rotation. — The  work 
done  by  any  force  upon  a  rotating  body  may  be  expressed  in  terms 
of  the  moment  of  the  force  and  the  angular  displacement  of  the  body. 

Let  A  (Fig.  193)  be  the  point  of  application  of  a  force,  O  being 
the  center  of  the  circle  described  by  A.  Let  P  =  magnitude  of 
force ;  r  =  length  OA  ;  L  =  moment  of  P  about  the  axis  of  rota- 
tion ;  (j>  =  angle  between  OA  and  P;  6  —  angle  between  OA  and 
some  fixed  line  in  the  plane  of  motion. 

*See,  however,  Art.  509. 


422  THEORETICAL   MECHANICS. 

If  the  body  receives  an  infinitesimal  angular  displacement  d6,  the 

displacement  of  A  is  rdO,  and  its  direction  is  perpendicular  to  OA. 

The  work  done  by  the   force   is  therefore 

dW=  Prsm<}>-dd. 

But  L  =  Pr  sin  <j> ; 

hence  dW=LdO. 

That  is,  for  an  infinitesimal  displacement,  the 
work  done  is  equal  to  the  product  of  the 
Fig.  193.  moment  of  the  force  into  the  angular  dis- 

placement. 
For  a  finite  displacement,  let  6'  and  6"  be  the  initial  and  final 
values  of  6  ;  then 


■/.: 


W=\      Ld6. 
J  0' 

If  L  remains  constant  during  the  displacement, 
W=L(d"  -  6'). 

509.  Equation  of  Work  and  Energy  in  Any  Plane  Motion. — 

Let  a  rigid  system,  restricted  to  plane  motion,  be  acted  upon  by  any 
forces,  and  receive  any  displacement  during  a  certain  interval.  Let 
M,  /,  /,  v,  (o,  <f>,  6  have  the  same  meanings  as  in  Arts.  443,  446 ;  let 
P  denote  the  vector  sum  of  the  applied  forces,  and  L  the  sum  of 
their  moments  about  the  mass-center. 

The  equation  of  motion  of  the  mass-center  is  (Art.  445) 

P=Mp, 

being  identical  with  the  equation  of  motion  of  a  particle  of  mass  M 
acted  upon  by  a  force  P.  Resolving  P  and  p  along  the  tangent  to 
the  path  described  by  the  mass-center,  and  integrating  the  resulting 
equation  exactly  as  in  Art.  353,  there  results  the  equation 

Wx=  \M{y"*-v'%  .         .         .     (6) 

in  which  v'  is  the  initial  and  v"  the  final  value  of  vy  and  Wl  is  the 
total  work  done  by  the  applied  forces  on  the  assumption  that  they 
act  at  the  mass-center. 

The  equation  of  angular  motion  is  (Art.  446) 

L  =  /(dcofdt). 


THEORY    OF    ENERGY.  423 

Multiplying   this   equation   member   by   member   by  the   equation 

dd  =  codt,    . 

LdQ  ===  lot  dco. 

a* 

Integrating,  f     Ldd  =  \I{w" 2  —  <o' 2), 

J e' 

in  which  co'  and  a>"  are  the  initial  and  final  values  of  a>,  and  0',  6"  the 
initial  and  final  values  of  6. 

The  first  member  of  this  equation  is  equal  to  the  work  which 
would  be  done  by  the  actual  forces  if  the  rotation  occurred  about  a 
fixed  axis  through  the  mass-center  (Art.  508).  Calling  this  quantity 
W 

W2  =  il(»"> -<*>'*).  .         .         .     (7) 

Referring  to  the  value  of  the  total  kinetic  energy  of  a  rigid  sys- 
tem, given  in  equation  (3),  it  is  seen  that  the  values  of  Wx  and  Wt 
given  by  equations  (6)  and  (7)  are  equal  respectively  to  the  incre- 
ments of  the  two  portions  of  the  kinetic  energy  which  have  been 
called  energy  of  translation  and  energy  of  rotation.  Hence  the 
following  important  principle  may  be  stated  : 

In  any  plane  motion  of  a  rigid  system^  the  increment  of  the  energy 
of  translation  is  equal  to  the  work  done  by  the  external  forces  com- 
puted as  if  they  were  applied  at  the  mass-center ;  the  increment  of 
the  rotational  energy  is  equal  to  the  work  done  by  the  external  forces 
computed  as  if  the  rotation  occurred  about  a  fixed  axis  through  the 
mass-center. 

Examples. 

1.  A  homogeneous  cylinder  of  mass  20  lbs.  and  radius  6  ins. 
rotates  about  its  axis  of  figure  at  the  rate  of  300  rev.  -per-min.  Re- 
quired the  value  of  the  kinetic  energy  in  foot-pounds. 

2.  The  mass  of  a  wheel-and-axle  is  50  lbs.,  its  radius  of  gyration 
with  respect  to  the  axis  of  rotation  is  10  ins.,  and  the  radius  of  the 
axle  is  6  ins.  It  is  set  rotating  by  a  constant  tension  of  5  pounds- 
force  in  the  rope  which  unwinds  from  the  axle.  Required  the  angular 
velocity  of  the  body  after  making  4  rev. ,  starting  from  rest. 

Ans.  1.72  rev.-per-sec. 

3.  Take  data  as  in  Ex.  2,  except  that  the  tension  is  due  to  a 
weight  of  5  lbs.  suspended  from  the  rope,  {a)  Determine  the  angular 
velocity  of  the  body  after  making  4  rev.  (b)  Determine  the  tension 
in  the  rope.     [Neglect  the  weight  of  the  rope  in  both  cases.  ] 

Ans.  (a)  1.69  rev.-per-sec.     (b)  4.91  lbs. 


424  THEORETICAL    MECHANICS. 

4.  A  body  of  any  shape  is  projected  upward  against  gravity  ; 
determine  the  height  to  which  the  mass-center  will  rise,  if  no  external 
force  acts  except  gravity.     Solve  by  the  principle  of  work  and  energy. 

[Notice  that  the  reasoning  by  which  equation  (6)  was  deduced  is 
valid  without  restricting  the  motion  to  a  plane.] 

5.  Apply  the  equation  of  work  and  energy  to  a  compound  pen- 
dulum, and  deduce  equation  (5)  of  Art.  425. 

6.  Solve  Ex.  9,  Art.  447,  by  the  principle  of  work  and  energy. 

7.  Solve  Ex.  10,  Art.  447,  by  the  principle  of  work  and  energy. 


§  5.    The  Principle  of  Virtual  Work. 

510.  Equation  of  Virtual  Work  for  a  System  of  Particles.— 
It  has  been  shown  (Art.  369)  that  if  a  particle  in  equilibrium  be 
assumed  to  receive  any  arbitrary  small  displacement,  the  total  work 
done  upon  it  by  all  forces  is  equal  to  zero.  By  simple  addition  this 
principle  may  be  extended  to  any  system  of  particles.     That  is, 

If  every  particle  of  a  system  is  in  equilibrium,  the  total  work 
done  upon  all  the  particles  during  any  arbitrary  small  displacements 
is  zero. 

In  general  the  ' '  equation  of  virtual  work ' '  obtained  by  applying 
this  principle  involves  both  internal  and  external  forces.  In  many 
important  cases,  however,  the  displacements  may  be  so  taken  as  to 
eliminate  certain  of  the  internal  forces. 

511.  Equation  of  Virtual  Work  for  Rigid  Body. —  Since  the 
total  internal  work  is  zero  for  a  displacement  which  leaves  the  con- 
figuration unchanged,  the  internal  forces  may  be  omitted  from  the 
equation  of  virtual  work  for  a  rigid  body.     That  is, 

If  a  rigid  body  is  in  equilibrium,  the  total  work  done  by  the  ex- 
ternal forces  during  any  arbitrary  displacement  is  zero. 

In  applying  this  principle,  it  is  generally  necessary  to  take  the 
displacement  infinitely  small,  since  a  finite  displacement  will  so 
change  the  relation  among  the  lines  of  action  of  the  forces  that  they 
will  no  longer  be  in  equilibrium. 

Since  the  negative  of  any  one  of  a  number  of  forces  in  equilibrium 
is  the  resultant  of  all  the  rest,  and  since  reversing  a  force  reverses  the 
sign  of  the  work  done  by  it  without  otherwise  changing  the  value,  it 
follows  that  the  total  quantity  of  work  done  by  any  number  of  ex- 
ternal forces  acting  upon  a  rigid  body  during  any  displacement  is 
equal  to  the  work  done  by  their  resultant. 


THEORY    OF    ENERGY.  425 

512.  Equations  of  Equilibrium  for  a  Rigid  Body. —  The  gen- 
eral equations  of  equilibrium  for  a  rigid  body,  already  deduced 
in  Part  I,  may  be  obtained  by  applying  the  principle  of  virtual 
work. 

Thus,  if  a  body  be  conceived  to  receive  an  arbitrary  small  trans- 
lation in  a  given  direction,  the  virtual  work  of  any  force  is  equal  to 
the  product  of  the  displacement  into  the  resolved  part  of  the  force 
in  the  direction  of  the  displacement ;  and  the  total  virtual  work  is 
equal  to  the  product  of  the  displacement  into  the  sum  of  the  re- 
solved parts  of  the  forces  in  the  direction  of  the  displacement.  Since 
this  total  work  is  zero,  whatever  the  direction  of  the  displacement, 
it  follows  that,  for  equilibrium,  the  algebraic  sum  of  the  resolved 
parts  of  the  forces  in  any  direction  is  zero. 

Again,  let  the  virtual  displacement  be  a  rotation  about  a  fixed 
axis  M.  The  work  done  by  any  force  is  equal  to  the  product  of  the 
angular  displacement  into  the  moment  of  the  force  about  M  (Art. 
508) ;  and  the  total  virtual  work  is  equal  to  the  angular  displacement 
into  the  sum  of  the  moments  of  the  forces  about  M.  Since  this 
total  work  is  zero,  whatever  the  position  of  the  axis  of  rotation,  it 
follows  that,  for  equilibrium,  the  sum  of  the  moments  of  the  forces 
about  any  axis  is  equal  to  zero. 

These  two  general  forms  include  all  the  equations  of  equilibrium 
for  forces  applied  to  a  rigid  body,  whether  in  two  dimensions  or  in 
three,  as  already  given  in  Chapter  X. 

513.  Connected  Bodies. —  If  the  principle  of  virtual  work  is  ap- 
plied to  a  system  of  rigid  bodies  connected  in  any  manner,  as  by 
hinges  or  strings,  or  pressing  against  one  another,  the  equation  of 
work  must  in  general  include  {a)  all  external  forces  acting  upon  any 
of  the  bodies  and  (b)  the  forces  exerted  by  the  bodies  upon  one 
another  by  reason  of  their  connections.  In  certain  important  cases, 
however,  forces  due  to  connections  may  be  omitted  from  the  equation 
because  it  can  be  seen  that  their  work  vanishes. 

Thus,  any  two  adjacent  portions  of  a  tense  string  exert  upon  each 
other  forces  constituting  a  tensile  stress.  If  the  distance  between 
the  two  portions  remains  constant,  the  total  work  of  the  stress  is 
zero  (Art.  482).  But  if  the  length  of  the  string  changes,  the  total 
work  of  the  internal  forces  will  not  be  zero.  If  a  system  includes 
two  bodies  which  are  connected  by  a  string,  the  tension  in  this  string 
will  not  enter  the  equation  of  virtual  work  for  the  system  if  the  dis- 


426  THEORETICAL    MECHANICS. 

placement  is  such  that  the  string  remains  tight  and  of  unchanged 
length. 

The  same  is  true  of  the  contact-stress  between  two  bodies  touch- 
ing each  other,  if  the  surfaces  of  contact  are  smooth  and  the  dis- 
placement is  such  as  not  to  destroy  the  contact.  Thus,  the  action 
and  reaction  between  two  bodies  connected  by  a  smooth  hinge  joint 
may  be  omitted  from  the  equation  of  virtual  work.  But  if  the  sur- 
faces are  rough  and  the  displacement  is  such  that  sliding  occurs,  the 
frictional  work  must  enter  the  equation. 

514.  Solution  of  Problems  in  Equilibrium. —  The  principle  of 
virtual  work  furnishes  a  simple  solution  of  many  problems  relating 
to  the  equilibrium  of  rigid  bodies,  or  of  connected  systems  of  bodies. 
The  method  of  applying  the  principle  is  to  assume  an  arbitrary  in- 
finitesimal displacement,  compute  the  work  done  by  all  the  applied 
forces,  and  equate  the  total  work  to  zero.  By  choosing  different 
displacements,  different  equations  may  be  obtained.  In  general,  for 
a  rigid  body  acted  upon  by  coplanar  forces,  three  independent  equa- 
tions may  be  formed. 

Any  force  may  be  eliminated  by  taking  the  displacement  of  its 
point  of  application  perpendicular  to  the  direction  of  the  force.  By 
this  means  two  external  forces  acting  on  a  rigid  body  may  be  elim- 
inated, since  the  directions  of  displacement  of  two  points  may  be 
chosen  arbitrarily. 

515.  Simple  Machines  Treated  by  Principle  of  Work.— The 
principle  of  work  often  furnishes  the  most  convenient  method  of  de- 
termining the  relation  between  effort  and  load  in  the  case  of  a  simple 
machine  or  a  combination  of  simple  machines. 

System  of  pulleys. —  To  determine  the  relation  between  effort 
and  load  in  case  of  a  system  of  pulleys  arranged  as  in  Fig.  1 94. 

The  system  consists  of  two  sets  of  pulleys,  each  set  comprising 
three  pulleys  mounted  freely  upon  a  common  axis  and  free  to  rotate 
independently  of  one  another ;  and  a  cord  passing  continuously 
around  the  pulleys  as  shown,  one  end  being  attached  at  A,  and  the 
effort  P  being  applied  at  the  other  end.  The  axis  of  the  set  A  is 
firmly  attached  to  a  fixed  support  C,  while  the  load  W  is  applied  to 
B.  This  * '  load ' '  may  be  the  weight  of  a  heavy  body,  or  it  may  be 
a  force  applied  horizontally  or  in  any  other  direction. 

Assuming  the  cord  to  be  perfectly  flexible  and  inextensible,  and 


THEORY   OF    ENERGY. 


427 


neglecting  axle  friction,  the  principle  of  work  may  be  applied  as 
follows : 

Let  the  system  be  displaced  in  such  a  way  as  to  shorten  the  dis- 
tance AB  by  an  amount  h,  the  cord  remaining  tense.     The  straight 
portions  of  the  cord  being  practically  parallel  to  AB, 
the  end  of  the  cord  moves  a  distance  6k.     The  effort 
P  and  the  resistance  W  are  the  only  forces  entering 
the  equation  of  work  for  the  system  composed  of  the 
two  sets  of  pulleys  and  the  cord,  and  we  have 
work  done  by  effort  =  6P/1 ; 
"      "  load    =  —  Wh. 


The  equation  of  work  is  therefore 

6Pk—  Wh  =  o;    .-.     P 


W/6. 


Train  of  toothed  wheels. — The  principle  of  work 
may  be  applied  without  difficulty  to  any  connected 
series  of  toothed  wheels,  one  of  which  is  acted  upon 
by  a  force  tending  to  produce  rotation,  and  another 
by  a  counterbalancing  force. 

Fig.   195  represents  three  pairs  of  wheels,  freely 
mounted  upon  fixed  axes  at  A,  B  and  C,  so  connected 
by  teeth  that  an  angular  displacement  of  one  must  be  accompanied 
by  a  proportional  angular  displacement  of  each  of  the  others.     A 

force  P,  applied  as  shown,  tends  to  pro- 
duce a  displacement  in  one  direction, 
while  Q  tends  to  produce  an  opposite  dis- 
placement. 

The  relation  between  the  displace- 
ments of  any  two  points  of  the  system 
may  readily  be  determined  from  the  di- 
mensions and  manner  of  connection.  Neglecting  axle  friction  and 
friction  at  the  surfaces  of  contact  of  the  teeth,  the  equation  of  work 
for  the  system  shown  will  contain  no  forces  except  P  and  Q.  If  a 
and  b  are  the  distances  moved  by  the  points  of  application  of  P  and 
Q  respectively,  the  equation  of  work  is 

Pa—  Qb  =  o. 

Pulleys  and  belts. —  Let  motion  be  transmitted  from  one  set  of 
machinery  to  another  through  a  pair  of  pulleys,  rigidly  connected 


Fig.  195. 


428  THEORETICAL    MECHANICS. 

and  mounted  freely  upon  a  common  axis,  each  connected  by  a  belt 
with  another  pulley  from  which  it  receives  (or  to  which  it  imparts) 

motion   by   means  of  the   belt.     (Fig. 
196.) 

Assuming  the  belts  to  be  perfectly 

flexible  and  inextensible,  and  neglecting 

*"■*        axle  friction,  the  equation  of  work  may 

be  written'  as  follows  : 

"**  Let  a  and  b  be  the  radii  of  the  two 

Fig.  196.  pulleys,   T  and    T'   the  tensions  in  the 

two  portions  of  one  belt,  S  and  S'  the 

tensions  in  the  two  portions  of  the  other.     Then,  for  any  angular 

displacement  A6,  the  equation  of  work  is 

TaAd  —  T'aAd  —  SbAO  +  S'bAO  =  o; 
or  (T—  T')a=(S  —  S')b. 

Screw. —  A  screw,  used  as  a  machine,  is  arranged  to  work  in  a 
fixed  nut,  and  is  usually  provided  with  an  arm  rigidly  attached  to  the 
screw  and  at  right  angles  to  its  length.  The  effort  is  applied  at  the 
end  of  this  arm  and  perpendicularly  to  it  and  to  the  axis  of  the 
screw,  while  the  load  is  applied  at  the  end  of  the  screw,  parallel  to  its 
length.  The  load  may  be  the  weight  of  a  heavy  body  which  is  to 
be  lifted,  the  fixed  nut  being  supported  upon  a  solid  foundation  in 
the  earth. 

Neglecting  friction,  the  relation  between  load  and  effort  may  be 
found  as  follows : 

Let  P  =  effort,  applied  at  distance  a  from  the  axis  of  rotation  of 
the  screw,  in  a  direction  perpendicular  to  the  arm  and  to  the  axis  of 
the  screw ;  W  =  load,  applied  in  a  direction  parallel  to  the  axis  of 
rotation  ;  b  =  pitch  of  screw  (distance  between  centers  of  two  con- 
secutive threads,  measured  parallel  to  the  axis  of  rotation). 

Let  the  screw  and  arm  rotate  at  a  uniform  rate,  so  that  the 
external  forces  are  in  equilibrium,  and  let  the  work  done  by  all  forces 
acting  upon  the  body  be  computed  for  one  revolution. 

The  work  done  by  P  is  2iraP ;  that  done  by  W  is  —  Wb.  The 
forces  exerted  by  the  fixed  body  do  no  work,  if  friction  is  neglected. 

The  total  work  done  upon  the  body  is 
2iraP  —  Wb  =  o  ; 
whence  P  ==  \Vb\21ra. 


THEORY    OF    ENERGY.  429 

By  making  b  small,  W  may  be  made  very  great  in  comparison 
with  P. 

The  following  examples  are  designed  to  illustrate  the  application 
of  the  principle  of  virtual  work.  They  are  all  problems  in  Statics, 
and  may  also  be  solved  by  the  methods  of  Part  I. 

Examples. 

1.  Determine  the  position  of  equilibrium  of  a  bar  resting  with 
both  ends  against  the  inner  surface  of  a  smooth  hemispherical  bowl. 

If  the  bar  be  displaced  from  the  position  of  equilibrium  in  such  a 
way  that  the  ends  slide  along  the  surface  of  the  bowl,  the  work  done 
by  the  reactions  is  zero  ;  hence  the  work  done  by  gravity  (the  only 
other  external  force)  must  be  zero.  This  requires  that  the  displace- 
ment of  the  center  of  gravity  shall  be  horizontal.  In  order  that  this 
requirement  shall  be  fulfilled  for  every  possible  sliding  displacement, 
the  center  of  gravity  must  be  vertically  below  the  center  of  the  bowl. 

2.  Find  the  position  of  equilibrium  of  a  heavy  bar  resting  with 
its  ends  upon  two  smooth  inclined  planes. 

Let  AB  (Fig.  197)  represent  the  bar,  C  being  its  center  of  gravity, 
and  let  it  receive  a  displacement  such  that  the  ends  slide  upon  the 
planes.  Reasoning  as  in  Ex.  1,  it  is  seen  that,  if  the  bar  is  displaced 
from  the  position  of  equilibrium,  the  displacement  of  C  must  be  hori- 
zontal if  the  ends  slide  along  the  planes,  since  the  work  done  by  the 
weight  of  the  bar  must  then  be  zero. 
From  this  condition  and  the  geometry 
of  the  figure,  the  position  of  equilibrium 
may  be  determined. 

Let  6  be  the  inclination  of  the  bar 
to  the  horizontal,  and  y  the  vertical  dis-    _  _  jSz^^/-@. 
tance  of  the  center  of  gravity  above  a  pIG#  I9y# 

horizontal  plane  through  the  line  of  in- 
tersection of  the  supporting  planes.     Let  a  and  /3  be  the  inclinations 
of  the  planes  to  the  horizontal,  A  C  =  a,  CB  =  b.     Let  y  be  ex- 
pressed as  a  function  of  6  ;  the  result  is 

y  =  {a  +  b) 21? sin  (fi  —  0)  +  a  sin  6. 

sin  (a  —  fi) 

As  the  bar  passes  through  the  position  of  equilibrium  while  sliding 
on  the  planes,  dy/dd  =  o. 

Differentiating  and  reducing, 

(a  -f-  b)  tan  6  =  a  cotan  a  —  b  cotan  ft. 

3.  A  bar  7  ins.  long,  whose  mass-center  is  3  ins.  from  one  end, 
rests  within  a  smooth  hemispherical  bowl  1 2  ins.  in  diameter.  What 
is  the  inclination  of  the  bar  to  the  vertical  when  in  equilibrium  ? 

Ans.  840  9. 


43o 


THEORETICAL    MECHANICS. 


4.  A  bar  1 2  ins.  long,  whose  mass-center  is  7  ins.  from  one  end, 
rests  upon  two  planes  inclined  300  to  the  horizontal.  Determine  the 
position  of  equilibrium.  Ans.  6  =  160  6'. 

5.  A  straight  bar  rests  with  one  end  upon  a  smooth  inclined 
plane  and  leans  against  a  smooth  peg.  Determine  the  position  of 
equilibrium. 

6.  A  bar  leans  against  a  smooth  peg,  the  lower  end  resting  upon 
a  smooth  surface.  Determine  the  form  of  the  surface  in  order  that 
the  bar  may  be  in  equilibrium  wherever  placed. 

An s.  A  section  of  the  surface  by  a  vertical  plane  is  represented 
by  the  polar  equation  r  =  a  -\-  c  sec  6. 

7.  A  uniform  bar  AB,  of  weight  G,  can  turn  freely  in  a  vertical 
plane  about  a  hinge  at  A .     To  B  is  attached  a  string  which  passes 

over  a  smooth  peg  at  C  and  sustains  a  weight  P. 
Determine  the  position  of  equilibrium. 

8.  A  screw  of  %  in.  pitch  is  used  to  raise  a 
weight  of  5,000  lbs.  Neglecting  friction,  what 
force  must  be  applied  at  the  end  of  an  arm  3  ft. 
long? 

9.  Determine,  by  the  principle  of  work  (neg- 
lecting friction),  the  relation  between  the  effort  (P) 
and  the  load  (  W)  in  case  of  the  differential  wheel- 
and-axle  (Fig.  198). 

Ans.  P/IV=  (d'  —  d)/2a. 

10.  In  Ex.  9,  let  the  radius  of  the  wheel  be  2  ft. 
and  the  radii  of  the  two  portions  of  the  axle  6  ins. 
and  4  ins.  If  the  load  is  1,000  lbs.,  what  must  be 
the  value  of  the  effort  ? 

11.  A  string  AB  is  attached  to  a  uniform  bar 
BC  of  which  the  end  C  rests  against  a  smooth  ver- 

Find  the  position  of  equilibrium. 
Ans.  If  the  inclinations  of  the  string  and  bar  to  the  vertical  are 
6  and  <j>  respectively,  tan  <j)  =  2  tan  0  in  the  position  of  equilibrium. 

12.  Two  uniform  bars,  AB,  BC,  of  unequal  lengths  and  masses, 
are  connected  by  a  smooth  hinge  at  B.  At  A  and  C  are  small  rings 
which  slide  on  a  smooth  horizontal  wire.  Determine  the  position  of 
equilibrium. 

Ans.  There  will  be  equilibrium  if  either  AB  or  B C  is  vertical. 
One  of  these  solutions  is  imaginary. 

13.  Take  data  as  in  Ex.  12,  except  that  the  bars  are  not  uniform, 
so  that  their  mass-centers  are  not  equally  distant  from  the  wire. 
Determine  the  position  of  equilibrium. 

14.  Two  equal  bars  connected  by  a  smooth  hinge  rest  upon  a 
smooth  cylinder  whose  axis  is  horizontal.  Determine  the  position 
of  equilibrium. 


*0 

Fig.  198. 
tical  plane  A  C. 


THEORY   OF    ENERGY.  43 1 

Ans.  Let  a  =  distance  of  mass-center  of  each  bar  from  hinge, 
r  =  radius  of  cylinder,  6  =  angle  between  either  bar  and  the  vertical 
in  the  position  of  equilibrium.     Then  a  sin30  —  r  cos  6  =  o. 

1 5.  Two  particles  of  unequal  masses,  connected  by  a  weightless 
inextensible  string,  rest  on  the  surface  of  a  smooth  cylinder  whose 
axis  is  horizontal.     Determine  the  position  of  equilibrium. 


CHAPTER  XXIV. 

RELATIVE    MOTION. 

§  i.  Meaning  of  Absolute  and  Relative  Motion. 

516.  Necessity  of  a  "Base"  for  Specifying  Motion.— Atten- 
tion has  been  called  (Art.  267)  to  the  fact  that,  in  order  to  specify 
the  motion  of  a  particle,  a  reference  body  must  be  chosen,  and  that 
the  values  of  the  velocity  and  the  acceleration  of  a  particle  estimated 
with  respect  to  one  body  are  in  general  different  from  their  values 
estimated  with  respect  to  another.  The  body  to  which,  in  any 
given  case,  the  motion  is  referred,  may  conveniently  be  called  the 
base. 

For  the  purposes  of  mathematical  analysis,  the  base  is  replaced 
by  certain  lines  forming  a  geometrical  frame.  Thus,  for  specifying 
motions  with  respect  to  the  earth,  a  frame  may  be  chosen  consisting 
of  three  rectangular  axes  intersecting  at  some  definite  point  on  the 
earth  and  having  directions  fixed  with  reference  to  the  earth. 

It  is,  however,  obvious  that  a  frame  need  not  be  fixed  to  any 
single  body  in  order  to  serve  as  a  base  for  specifying  motions.  For 
example,  the  directions  of  the  lines  forming  the  frame  may  be  speci- 
fied with  respect  to  the  fixed  stars. 

517.  "  True  "  and  "  Apparent  "  Motion. — When  the  motions 
of  bodies  near  the  earth  are  observed,  it  is  with  reference  to  the 
earth  that  those  motions  are  commonly  specified.  But  in  certain 
cases  the  rotation  of  the  earth  about  its  axis  is  brought  into  consid- 
eration, and  we  say  that  the  motion  of  a  body  referred  to  the  earth 
as  base  is  only  its  ' '  apparent ' '  motion  ;  and  that  the  motion  of 
the  earth  itself  must  be  taken  into  account  if  it  is  desired  to  deter- 
mine the  "true"  motion  of  the  body. 

It  is,  however,  obvious  that  in  taking  account  of  the  earth's  rota- 
tion what  we  are  really  doing  is  to  refer  the  motions  of  terrestrial 
bodies  to  a  new  base.  And  it  is  necessary  to  raise  the  question 
whether  there  is  any  reason  for  regarding  the  motion  referred  to 
this  new  base  as  the  true  motion,  or  even  as  being  a  nearer  approxi- 
mation to  the  true  motion  than  is  obtained  by  taking  the  earth  as 


RELATIVE   MOTION.  433 

base.  Is  it,  in  fact,  possible  to  assign  any  meaning  to  the  "real" 
motion  of  a  body  as  distinguished  from  its  "apparent"  motion? 

It  is  evident  that,  so  long  as  our  object  is  to  study  the  motions 
of  bodies  kinematically \  any  base  whatever  may  serve  the  purpose. 
What  base  shall  be  chosen  in  any  particular  case  is  purely  a  matter 
of  convenience.  From  this  point  of  view  there  is  no  reason  for 
drawing  a  distinction  between  true  and  apparent  motions.  Or,  as 
certain  writers  express  it,  one  base  gives  as  true  a  description  of  the 
motion  as  another,  although  simplicity  may  be  gained  by  choosing 
the  base  in  a  particular  way. 

But  when  motions  are  studied  causally \  it  becomes  evident  that 
the  choice  of  base  is  not  merely  a  matter  of  convenience.  It  will,  in 
fact,  be  seen  that  the  ' '  laws  of  motion ' '  cannot  be  true  independ- 
ently of  the  base  to  which  motion  is  referred. 

518.  Importance  of  Choice  of  Base  in  Applying  the  Laws  of 
Motion.  —  That  the  laws  of  motion,  if  true  for  one  base,  cannot  be 
true  for  another  which  moves  in  an  arbitrary  manner  with  respect  to 
the  first,  may  be  shown  by  a  simple  illustration. 

Conceive  a  horizontal  platform  to  be  rotating  uniformly  about  a 
vertical  axis  fixed  in  the  earth,  and  consider  the  motion  of  a  ball 
which  is  placed  upon  the  platform.  Let  this  motion  be  specified 
both  with  respect  to  the  platform  and  with  respect  to  the  earth,  and 
let  the  two  motions  be  compared.  *     , 

Suppose  first  that  the  ball  is  stationary  with  respect  to  the  plat- 
form. If  the  platform  be  taken  as  base,  the  acceleration  is  zero; 
and  if  the  laws  of  motion  be  applied,  the  inference  is  that  the  result- 
ant of  all  forces  acting  upon  the  ball  is  zero,  since  the  acceleration  is 
always  proportional  to  the  resultant  force.  But  if  the  earth  be  taken 
as  base,  the  body  describes  a  circle  at  a  uniform  speed.     If  a>  is  the 

*A  concrete  idea  of  what  is  meant  by  the  motion  with  respect  to  the 
platform  may  be  obtained  by  considering  how  the  motion  will  appear  to  an 
observer  stationed  upon  the  platform,  whose  view  is  so  obstructed  that  the 
earth  and  the  bodies  fixed  upon  it  are  invisible  to  him.  To  his  experience 
the  platform  is  a  "  fixed  "  body,  and  is  the  only  base  available  for  estimating 
the  motion  of  the  ball.  On  the  other  hand,  if  a  person  stationed  upon  the 
earth  can  observe  the  ball,  he  may  take  the  earth  as  his  base  for  estimating 
its  motion.  The  path  described  by  the  ball  will  appear  quite  different  to  the 
two  observers,  and  its  velocity  and  acceleration  as  estimated  by  them  will 
have  quite  different  values. 


434  THEORETICAL   MECHANICS. 

angular  velocity  of  the  platform  and  r  the  distance  of  the  body  from 
the  axis  of  rotation,  it  has  an  acceleration  with  respect  to  the  earth 
equal  to  o>V  directed  toward  the  axis.     The  general  equation 

force  =  mass  X  acceleration 

therefore  shows  that  the  resultant  of  all  forces  acting  upon  the  body 
is  a  force  of  magnitude  mroo?  directed  toward  the  center  of  the  cir- 
cular path;  m  being  the  mass  of  the  ball. 

As  another  case,  suppose  the  path  of  the  ball  upon  the  platform  is 
a  straight  line  passing  through  the  axis  of  rotation,  and  that  the 
velocity  estimated  with  respect  to  the  platform  is  constant.  If  the 
laws  of  motion  hold  when  the  platform  is  base,  the  resultant  force 
acting  upon  the  ball  must  be  zero,  since  its  acceleration  is  zero.  But 
with  respect  to  the  earth  the  body  describes  a  spiral  with  varying 
speed  ;  hence  if  the  laws  of  motion  be  applied  with  the  earth  as  base, 
the  inference  is  that  the  resultant  of  all  forces  acting  upon  the  ball 
varies  both  in  magnitude  and  in  direction. 

The  actual  forces  acting  upon  the  body  are,  however,  the  same 
whatever  be  the  base  of  reference.  Our  conception  of  a  force  is  that 
it  is  an  action  of  one  body  upon  another  ;  *  and  the  actions  of  other 
bodies  upon  the  ball  at  any  given  instant  cannot  be  supposed  to 
depend  upon  whether  the  motion  is  estimated  with  respect  to  the 
earth  or  to  the  platform. 

It  is  thus  evident  that  if  the  laws  of  motion  are  true,  and  if  our 
definition  of  force  (which  forms  an  essential  element  in  the  interpre- 
tation of  those  laws)  is  to  be  retained,  it  is  not  permissible  to  refer 
motions  to  a  base  chosen  arbitrarily  in  applying  the  laws.  It  is 
necessary,  in  fact,  to  raise  the  fundamental  question  whether  there  is 
any  base  for  which  the  laws  of  motion  are  true,  f 

519.  Ultimate  Base. —  It  is  obvious  that,  unless  there  is  a  base 
for  which  the  laws  of  motion  are  true,  these  laws  are  unintelligible.  % 

*See  Art.  212. 

tit  may  be  well  to  remark  that  this  is  not  the  same  as  the  question 
whether  such  a  base  can  be  exactly  determined  by  experiment. 

%  Maxwell,  in  his  discussion  of  Newton's  first  law  (  "Matter  and  Motion," 
Chapter  III),  argued  that  no  alternative  law  can  be  stated  which  is  intelligible 
"unless  we  admit  the  possibility  of  defining  absolute  rest  and  absolute  veloc- 
ity." And  denying  this  possibility,  he  appeared  to  regard  this  as  a  priori 
evidence  of  the  validity  of  the  law.     It  would  seem,  however,  that  the  argu- 


RELATIVE   MOTION.  435 

That  such  a  base  is  a  reality  must  in  fact  be  regarded  as  an  essential 
part  of  the  laws  themselves. 

Definition.  —  A  body  or  frame  of  reference  for  which  the  laws  of 
motion  are  true  may  be  called  an  ultimate  base. 

520.  Time.  —  In  order  to  specify  motion  it  is  necessary  to 
adopt  some  scale  for  estimating  time.  That  is,  some  means  must 
be  adopted  for  comparing  different  intervals  of  time. 

The  first  essential  is  to  be  able  to  specify  successive  equal  inter- 
vals of  time  ;  and  for  this  purpose  the  only  method  practically  avail- 
able is  to  observe  some  body  which  is  supposed  to  move  uniformly, 
or  to  perform  a  definite  series  of  motions  in  uniformly  recurring 
cycles.  Thus,  we  may  take  as  equal  the  successive  intervals  occu- 
pied by  complete  revolutions  of  the  earth  upon  its  axis  ;  or  the  inter- 
vals occupied  by  successive  vibrations  of  a  pendulum. 

Now  the  time-scale  specified  by  the  motion  of  one  body  will  not 
in  general  agree  exactly  with  that  specified  by  the  motion  of  another, 
and  the  question  arises  which,  if  any,  of  the  scales  thus  denned  is 
the  "true"  time-scale. 

So  far  as  a  mere  description  of  the  motions  of  bodies  is  concerned, 
any  time-scale  will  serve.  The  actual  values  of  the  velocity  and 
acceleration  of  a  particle  will  depend  upon  what  scale  is  adopted,  but 
a  time-scale  having  been  chosen,  all  motions  can  be  described  in 
terms  of  it.  Some  writers,  in  fact,  assert  that  one  time-scale  gives 
as  correct  a  description  of  motions  as  another,  and  that  the  only 
reason  for  preferring  one  to  another  is  a  gain  of  simplicity. 

It  is  obvious,  however,  that  the  laws  of  motion  cannot  be  true 
independently  of  the  time-scale.  Thus,  the  first  law  asserts  that  a 
body  not  acted  upon  by  force  (i.  e. ,  not  influenced  by  other  bodies) 
would  move  uniformly  in  a  straight  line.  But  motion  which  is  uni- 
form when  one  time-scale  is  used  may  not  be  uniform  when  another 
is  used. 

The  laws  of  motion  are,  in  fact,  wholly  unintelligible  unless  we 

ment  which  Maxwell  applied  to  the  negation  of  the  law  can  be  applied  with 
equal  validity  to  its  affirmation.  If  the  reasoning  is  valid,  the  conclusion 
appears  to  be  that  the  laws  of  motion  as  stated  by  Newton  are  unintelligible. 
That  Newton's  conception  of  absolute  motion  corresponds  to  a  reality 
can  hardly  be  denied,  unless  we  are  prepared  to  discard  the  whole  New- 
tonian system.  Whether  the  term  "absolute"  is  properly  applied  is 
another  question. 


436  THEORETICAL   MECHANICS. 

admit  the  reality  of  a  time-scale  for  which  they  are  true ;  and  this 
reality  must  be  accepted  as  a  part  of  the  laws  themselves.  * 

521.  Absolute  Space  and  Time. —The  reality  of  an  ultimate 
base  for  specifying  motion,  and  of  an  ultimate  scale  for  estimating 
time,  was  explicitly  affirmed  by  Newton.  He  in  fact  postulated 
absolute  space  and  absolute  time  as  the  space  and  time  implied  in 
his  statement  of  the  laws  of  motion. 

Many  modern  writers  have  found  the  Newtonian  conception  of 
absolute  space  and  time  so  repugnant  that  they  have  discarded  it. 
The  attempt  to  destroy  the  force  of  Newton's  arguments  has  not, 
however,  been  successful,  and  as  already  stated,  there  appears  to  be 
as  much  reason  for  accepting  his  conception  of  absolute  motion  as 
corresponding  to  reality  as  for  accepting  the  Newtonian  laws  as  a 
sound  basis  for  the  science  of  Mechanics. 

522.  Absolute  and  Relative  Motion.— The  motion  of  a  body 
with  respect  to  an  ultimate  base  may  be  defined  as  its  absolute  mo- 
tion ;  and  its  motion  referred  to  any  other  than  an  ultimate  base 
may  be  called  its  relative  motion,  f 

In  certain  cases  it  is  necessary  to  compare  the  motions  of  a  par- 
ticle as  estimated  with  respect  to  different  bases.  In  such  cases  it  is 
convenient  to  call  one  of  the  motions  absolute,  even  though  the  base 
of  reference  is  not  an  ultimate  one.  Thus,  there  are  many  practical 
problems  in  which  the  motion  of  a  particle  with  reference  to  the 
earth  has  to  be  compared  with  its  motion  referred  to  some  body 
which  is  itself  in  motion  with  respect  to  the  earth.  The  theory  of 
turbine  water-motors  presents  such  a  case,  it  being  needful  to  esti- 
mate the  motion  of  the  particles  of  water  both  with  respect  to  the 
rotating  wheel  and  with  respect  to  the  earth.  The  latter  may  be 
called  the  absolute  motion  to  distinguish  it  from  the  former.  In 
such  practical  problems  the  earth  is,  in  fact,  the  ultimate  base  to 
which  we  refer  motions,  although  it  is  known  not  to  be  a  true  ulti- 
mate base  as  above  defined.  The  error  which  results  from  regard- 
ing the  earth  as  an  ultimate  base  is  for  many  purposes  unimportant.  \ 

*It  may  be  well  to  remark  that  to  affirm  the  reality  of  a  time-scale  which 
is  independent  of  the  motions  of  actual  bodies  is  not  to  affirm  that  we  can 
practically  attain  to  such  a  scale. 

t  However  objectionable  the  word  ' '  absolute ' '  may  be  in  itself,  its  em- 
ployment seems  to  be  justified  by  usage. 

X  A  more  definite  discussion  of  the  earth  as  a  base  is  given  in  Art.  531. 


RELATIVE    MOTION.  437 

§  2.   Transformation  from  One  Base  to  Another. 

523.  Change  of  Base.  —  In  the  foregoing  discussion  the  impor- 
tance of  the  choice  of  a  base  for  specifying  the  motion  of  a  particle 
has  been  shown  by  simple  illustrations,  but  the  influence  of  the 
choice  of  base  upon  the  values  of  the  velocity  and  acceleration  has 
not  been  explained  in  an  exact  manner.  A  mathematical  discussion 
will  now  be  given,  showing  how  the  values  of  the  quantities  which 
specify  the  motion  are  changed  by  changing  from  one  base  of  refer- 
ence to  another  having  a  given  motion  with  respect  to  the  first. 

To  discuss  the  problem  in  its  full  generality  would  be  beyond 
the  scope  of  this  book,  since  the  treatment  already  given  of  the  mo- 
tion of  a  particle  and  of  a  rigid  body  does  not  go  beyond  plane  mo- 
tion and  motion  of  translation.  The  following  discussion  is  there- 
fore restricted  in  the  same  manner. 

The  case  of  translatory  motion  will  be  considered  first ;  then  a 
restricted  case  of  plane  motion  (that  in  which  the  relative  motion  of 
the  bases  is  a  uniform  rotation)  ;  and  finally  the  general  case  of 
plane  motion. 

Of  the  two  bases  or  frames  to  which  the  discussion  refers,  one 
may  be  called  the  primary,  the  other  the  secondary  ;  and  the  motion 
referred  to  the  former  will  often  be  called  the  absolute  motion,  while 
that  referred  to  the  latter  is  called  the  relative  motion. 

524.  Case  in  which  the  Relative  Motion  of  the  Bases  is 
Translatory.  —  Let  the  motion  of  a  particle  be  referred  to  a  base 
M\  and  also  to  a  base  M  which  itself  has  any  motion  of  translation 
with  respect  to  M' .  Let  x ',  y\  z'  be  the  coordinates  of  the  particle 
referred  to  a  set  of  rectangular  axes  0'X\  O'Y',  O'Z',  fixed  in 
M' ;  x,  y,  z  its  coordinates  with  respect  to  axes  OX,  O  Y,  OZ,  fixed 
in  M;  and  x0,  y0,  zQ  the  coordinates  of  the  origin  O  referred  to  the 
axes  OX',  O'Y',  O'Z'. 

The  position,  velocity  and  acceleration  of  the  particle  with  respect 
to  M'  are  given  by  the  values  of  x ,  y ,  z ,  and  their  first  and  second 
derivatives  with  respect  to  the  time  ;  the  position,  velocity  and 
acceleration  with  respect  to  M  by  the  values  of  x,  y,  z,  and  their 
first  and  second  derivatives.  Also,  x0,  y0,  z0,  and  their  derivatives 
specify  the  position,  velocity  and  acceleration  of  .^f  relative  to  M' . 


438  THEORETICAL    MECHANICS. 

The   relation  between  these  different  motions  is  expressed  by  the 
following  equations  : 

x*  =  x%  +  x,  \ 

y'  =io  +  y>  >     .      .      .      .    (i) 

z'  =  z0  -f  z;  ) 


CO 


x' 

=  x0  +  X, 

y 

=  >o  +  >> 

z 

=  z0  +  z  ; 

x' 

==  -^o  ~i     X, 

y 

=  yo  +  j;> 

r 

=  4  -f-  * ; 

(3) 


Equations  (2)  express  the  proposition  that 

The  velocity  of  the  particle  with  respect  to  the  base  M'  is  equal 
to  the  vector  sum  of  its  velocity  with  respect  to  M  and  the  velocity  of 
M  with  respect  to  M '. 

Similarly,  equations  (3)  state  that 

The  acceleration  of  a  particle  with  respect  to  the  base  M'  is  equal 
to  the  vector  sum  of  its  acceleration  with  respect  to  M  and  the  accel- 
eration ofM  with  respect  to  M '. 

Special  case. — If  M  has  a  uniform  rectilinear  motion  with  respect 
to  M ',  x0 ,  y0 ,  and  z0  are  all  zero.  In  this  case  the  acceleration  of  a 
particle  with  respect  to  M'  is  equal  to  its  acceleration  with  respect 
to  M. 

Examples. 

1.  A  railway  train  moves  uniformly  on  a  straight  track  at  the 
rate  of  25  miles  per  hour.  A  passenger  throws  a  stone  so  that  to 
him  it  appears  to  move  at  right  angles  to  the  track  with  a  velocity 

r  20  ft.  -per-sec.     What  is  its  velocity  with  respect  to  the  earth  ? 

2.  In  Ex.  1,  how  must  the  stone  be  thrown  in  order  that  its 
velocity  referred  to  the  earth  shall  be  20  ft. -per-sec.  at  right  angles 
to  the  track  ? 

3.  A  railway  train  moves  along  a  straight  track  in  such  a  way 
that  in  haif  a  minute  its  speed  increases  uniformly  from  10  miles  per 
hour  to  35  miles  per  hour.  A  stone  dropped  from  a  car  window 
has  what  acceleration  relative  to  the  earth  while  falling  to  the 
ground  ?     What  acceleration  has  it  relative  to  the  train  ? 

4.  In  Ex.  3,  compute  the  velocity  of  the  stone  relative  to  the 
train  after  it  has  fallen  4  ft.  vertically. 


RELATIVE    MOTION.  439 

525.  Case  in  which  the  Relative  Motion  of  the  Bases  is  a  Uni- 
form Rotation.* — Let  the  motion  of  a  particle  be  referred  to  a  rigid 
body  or  base  M',  and  also  to  a  body  M  which  rotates  uniformly 
about  an  axis  fixed  in  M' .  It  will  be  convenient  to  speak  of  M'  as 
a  ' '  fixed ' '  body,  the  motion  of  the  particle  relative  to  it  being  called 
the  "absolute"  motion,  and  that  relative  to  M  the  "relative" 
motion. 

Let  r  denote  the  distance  of  the  particle  from  the  axis  of  rotation 
at  time  t ;  ft)  the  angular  velocity  of  the  rotation  of  M  relative  to 
M' ;  vy  p  the  relative  velocity  and  acceleration  ;  v'f  p'  the  abso- 
lute velocity  and  acceleration.  It  is  required  to  find  (a)  the  relation 
between  v  and  v\  and  (^)  the  relation  between  p  and  p' .  It  will 
at  first  be  assumed  that  the  particle  moves  in  a  plane  perpendicular 
to  the  axis  of  rotation. 

(a)  Relation  between  velocities. — Let  Fig.  199  represent  the 
plane  of  the  motion,  O  being  the  point  in  which  this  plane  is  pierced 
by  the  axis  of  rotation.  Let  AB  be  the  path  traced  by  the  particle 
on  the  rotating  body  during  a  short  interval  At.  During  that  time 
this  body  turns  through  an  angle  coAt,  and  AB  is  carried  into  a 
position  A'B'  such  that  the  angles  AOA  '  and  BOB'  are  each  equal 
to  coAt.  The  path  traced  by  the  particle  upon  the  fixed  body  M'  is 
s6me  curve  AB' . 

The  velocity  of  the  particle  relative  to  the  rotating  body  at  the 
beginning  of  the  interval  At  has  the  direction  of  the  tangent  to  AB 
at  Ay  and  the  magnitude 

V  —  lim  [(vector  AB)/At]. 

The  absolute  velocity  at  the  same  instant  has  the  direction  of  the 
tangent  to  AB'  at  A,  and  the  magnitude 

v*  =  lim  [(vector  AB')IAt]. 

But  vector  AB'  =  vector  AA  '  +  vector  A'B' ; 

hence      v'  =  lim  [(vector  A'B')/ At]  +  lim  [(vector  AA')/At]. 

*  This  case  is  a  special  case  of  that  treated  in  Art.  526.  The  geomet- 
rical discussion  here  given  has  the  advantage  of  showing  more  vividly  the 
relations  of  the  vector  quantities  involved. 


44Q 


THEORETICAL    MECHANICS. 


If  u  denotes  the  velocity  of  the  point  A  of  the  body  M  due  to  the 
rotation, 

lim  [(vector  A  A  ')/At]  =  u. 

(The  direction  of  u  is  perpendicular  to  AO  and  its  magnitude 
is  rco.)  Also,  in  passing  to  the  limit  {At  approaching  o),  vector  AB 
may  replace  vector  A'B'.     Hence 

vector  v'  =  vector  v  -\-  vector  u. 
That  is, 

The  absolute  velocity  of  the  particle  is  equal  to  the  vector  sum  of 
its  relative  velocity  and  the  velocity  of  that  point  of  the  rotating 
body  which  is  the  instantaneous  position  of  the  particle. 


Fig.  199. 

ib)  Relation  between  accelerations. —  In  order  to  compare  the  ab- 
solute and  relative  accelerations,  let  the  value  of  each  be  computed 
in  the  manner  explained  in  Art.  250. 

Let  Av  denote  the  increment  of  the  relative  velocity  v  during  the 
time  At,  and  Av'  the  increment  of  v'  during  that  time.     Then 

/  =  lim  {AvjAt)  ; 

p'  =  lim  {Av'/At). 

We  may  first  compare  Av  and  Av '. 


RELATIVE   MOTION.  44 1 

The  value  of  Av  is  found  by  drawing  from  some  point,  Q,  vectors 
to  represent  the  initial  and  final  values  of  v.  The  former  is  parallel 
to  the  tangent  to  AB  at  A  and  is  represented  by  QA"  (  Fig.  199  )  ; 
the  latter  is  parallel  to  the  tangent  to  AB  at  B*  and  is  represented 

by  QB  "  ;  so  that 

vector  A" B"  =  Av. 

To  find  the  value  of  Av '  we  must  in  like  manner  draw  from  some 
point  vectors  representing  the  initial  and  final  values  ofv'.  These 
are  parallel  to  the  tangents  to  the  absolute  path  AB '  at  A  and  at  B ' 
respectively.  Having  drawn  QA"  to  represent  the  initial  value  of 
v,  that  of  v'  is  found  by  combining  with  QA"  a  vector  RQ  of  magni- 
tude rco  directed  at  right  angles  to  OA  (representing  the  velocity  u 
of  the  point  A  of  the  rotating  body).  This  gives  RA"  as  the  vec- 
tor value  of  the  velocity  v '  at  the  beginning  of  At.  Similarly,  the 
final  value  is  found  as  the  sum  of  two  components  representing  values 
of  v  and  u  at  the  end  of  At.  We  have  already  drawn  QB"  to  repre- 
sent the  final  value  of  v  ;  but  since  we  are  now  referring  velocities  to 
the  fixed  body,  the  direction  of  this  vector  must  be  made  parallel  to 
the  tangent  to  A'B'  at  B'.  That  is,  we  take  a  vector  QT equal  in 
magnitude  to  QB"  but  making  with  it  an  angle  co  At  (the  angle 
turned  through  in  time  At  by  the  tangent  to  AB  at  B  ).  The  other 
component  of  the  final  value  of  v '  is  perpendicular  to  OB '  and  of 
magnitude  OB'-  co  or  r1co.  Represent  this  latter  component  by  RS, 
and  draw  SB'"  equal  to  vector  QT;  then  RB'"  represents  the  final 
value  of  v ',  and 

vector  A" B'"   =  Av'. 

It  remains  to  compare  Av  and  Av '. 

The  following  vector  equation  may  be  written: 

Av'  =  Av  +  B"T  +  TB'"  =  Av  +  B"T  +  QS. 

Also  QS  may  be  expressed  as  the  sum  of  two  component  vectors 
QH,  HSy  the  point  H  being  located  as  follows  :  Make  RH  equal 
in  length  to  RQ  and  angle  QRH  =  coAt ;  then  RH  is  perpen- 
dicular to  OA  '.     This  construction  makes  the  triangle  RHS  similar 

*  Although  the  path  AB  is  at  A'B'  at  the  end  of  the  interval,  QB"  is 
drawn  parallel  to  the  tangent  at  B  in  the  original  position,  because  when  the 
rotating  body  is  taken  as  base  the  relation  between  initial  and  final  velocities 
is  not  affected  by  the  rotation.  The  angle  between  these  velocities  is  the 
angle  between  the  tangents  to  AB  at  A  and  B. 


442  THEORETICAL    MECHANICS. 

to  OA' B' ,   the    corresponding    sides  being  perpendicular  each    to 
each. 

Replacing  QS  by  QH  -j-  HS  in  the  last  equation,  dividing 
through  by  At  and  passing  to  the  limit,  At  approaching  o, 

p'  =/  +  lim[C£T)/A*]  +  lim  [(QH)/At] 

+  Um  [GffS)/A/].  (2) 

It  is  easy  to  evaluate  the  last  three  terms.  Consider  them  in  order, 
(i)  As  At  approaches  o,  the  angle  B"QT  approaches  o  and 
QB "T  approaches  a  right  angle,  while  QB  "  approaches  the  direction 
QA".  Hence  the  limiting  direction  of  the  vector  B"T\s  perpen- 
dicular to  QA",  that  is,  perpendicular  to  the  relative  velocity  v  at 
the  time  /.  The  limiting  magnitude  of  (B"T)jAt  may  be  found 
by  considering  the  circular  arc  B'T  whose  center  is  Q.     Thus 

lim  [(chord  B"T)/At]  =  lim  [(arc  B"T)\At\ 
=  lim  [(^o)  At)/At] 
=  lim  \t.\<*>\  =  vco. 

[  vx  is  written  for  the  value  of  v  at  the  end  of  At.  ] 

(2)  In  a  precisely  similar  manner  the  limiting  direction  and  mag- 
nitude of  (vector  QH)/At  may  be  found  from  the  triangle  QRH. 
Its  limiting  direction  is  perpendicular  to  RQ,  i.  e.,  it  is  AO.  Its 
limiting  magnitude  is 

RQa—  rooco  =  rco2. 

(3)  The  vector  US  is  perpendicular  to  the  chord  A'B'.  As 
At  approaches  o,  A'B'  approaches  AB,  while  chord  AB  ap- 
proaches tangent  to  curve  AB  at  A.  Hence  the  limiting  direction 
of  HS  is  perpendicular  to  the  relative  velocity  v.  The  limiting 
magnitude  of  (HS)/At  is  found  by  comparing  the  similar  triangles 
RHSy  OA'B'.  Each  side  of  the  former  triangle  is  equal  to  a) 
times  the  corresponding  side  of  the  latter.      That  is, 

lim  [(//5)/A/]  =  lim  [(v A'B')/ At]  =  <o  lim  [(A'B')/At]  =  cov. 

Noticing  that  the  first  and  third  of  the  three  vector  quantities 
just  evaluated  are  equal,  it  is  seen  that  the  last  three  terms  of  equa- 
tion (2)  are  equivalent  to  the  following  two  vector  quantities: 

An  acceleration  of  magnitude  r  co'2  directed  along  the  perpendicu- 
lar drawn  from  the  particle  to  the  axis  of  rotation. 

An  acceleration  2cov  directed  at  right  angles  to  the  relative  velocity. 


RELATIVE    MOTION.  443 

The  vector  equation  (2),  expressing  the  relation  between/ and 
p ',  may  be  written 

/'—/+  IXI  +  l2VCOl     .  •       •      •    (3) 

the  quantities  in  brackets  being  the  magnitudes  of  the  two  vector 
quantities,  and  their  directions  being  as  just  stated.  It  should  be 
stated  further  that  the  direction  of  the  component  [22/0)]  is  that 
obtained  by  rotating  the  vector  representing  v  through  a  right  angle 
in  the  direction  in  which  the  rotating  body  turns. 

Three-dimensional  motion  of  a  particle. — If  the  particle  is  not 
restricted  to  a  plane  perpendicular  to  the  axis  of  rotation,  Fig.  199 
may  still  represent  the  projection  of  its  motion  upon  such  a  plane. 
The  propositions  just  deduced  express  the  relation  between  the  re- 
solved parts  of  velocities  and  accelerations  parallel  to  this  plane. 
Thus,  if  a,  a  denote  the  angles  made  by  v,  v'  with  the  axis  of  rota- 
tion, and  /3,  ft'  the  angles  made  by p,  p'  with  that  axis,  we  may 
write  instead  of  equations  (1)  and  (3)  the  vector  equations 

[^'sin  a']  =  [y  sin  a]  -j-  u  ;      .  .  .  .     (1') 

[/'siny8']  =  [p  sin  /3]  -f-  [roo*]  -f-  \_2(ov  sin  a].      (3') 

Now  the  resolved  part  of  v  parallel  to  the  axis  is  equal  to  that  of 
V$\  and  by  adding  these  equal  vectors  to  the  two  members  of  (1')  we 
have 

v'  =v  -\-  u (1") 

Similarly,  since  the  resolved  parts  ofp  and/'  parallel  to  the  axis  are 
equal,  we  get  from  (3') 

/'  =  p  -\-  [rco2]  -j-  [2cov  sin  a].  .     (3") 

Equations  (1")  and  (3")  replace  (1)  and  (3)  when  the  motion  of  the 
particle  is  unrestricted.  It  is  seen  that  (1")  is  identical  with  (1), 
while  (3")  differs  from  (3)  only  in  the  last  term  ;  this  component  of 
acceleration  is  always  perpendicular  to  the  axis  of  rotation,  and  is 
computed  from  the  resolved  part  of  v  perpendicular  to  that  axis. 

Examples. 

1.  A  horizontal  platform  rotates  uniformly  about  a  vertical  axis 
fixed  in  the  earth,  at  the  rate  of  100  rev.-per-min.  A  particle  at 
rest  on  the  platform  5  ft.  from  the  axis  of  rotation  has  what  velocity 
and  what  acceleration  relative  to  the  earth  ? 

Ans.  The  magnitudes  of  the  velocity  and  the  acceleration  are 
v'  =  52.4  ft.-per-sec.,/'  =  548ft.-per-sec.-per-sec. 


444  THEORETICAL    MECHANICS. 

2.  A  horizontal  platform  rotates  with  uniform  angular  velocity  co 
about  a  vertical  axis  fixed  in  the  earth.  A  particle  describes  upon  it 
a  straight  path  passing  through  the  axis,  with  uniform  relative  velo- 
city v.  Determine  the  velocity  and  the  acceleration  relative  to  the 
earth  when  the  particle  is  distant  r  from  the  axis. 

3.  In  Ex.  2  let  a)  =  20  rev.  per  min.,  v  =  12  ft.  per  sec.  Solve 
for  O)  r  =  o,  (ti)  r  =  5  ft. 

Ans.  {b)  v'  =  15.9  ft.  per  sec,  inclined  1380  54'  to  perpendicu- 
lar from  particle  upon  axis  ;  p'  =  54.8  ft.-per-sec.-per-sec,  at  angle 
66°  26'  to  same  line. 

4.  In  Ex.  2,  let  the  shortest  distance  of  the  path  from  the  axis 
be  /*,  the  other  data  being  unchanged. 

5.  Assuming  the  platform  to  rotate  as  in  Ex.  3,  suppose  a  body 
to  fall  freely  from  rest  under  gravity  from  a  point  8  ft.  from  the  axis 
of  rotation.  Determine  its  velocity  and  acceleration  relative  to  the 
platform  after  falling  o.  5  sec. 

526.  Case  in  which  the  Relative  Motion  of  the  Bases  is  Any- 
Plane  Motion. — Let  the  motion  of  either  base  relative  to  the  other 
be  any  plane  motion,  and  let  the  particle  be  restricted  to  a  plane 
common  to  the  two  bases,  represented  in  Fig.  200.  Let  x,  y  be  the 
coordinates  of  the  particle  referred  to  axes  OX,  O  Fin  the  secondary 
base  M\x' ,  y\  its  coordinates  referred  to  axes  O X,'  O'Y'  in  the 
primary  base  M';  xw  y0  the  coordinates  of  the  origin  O  referred 
to  OX\  O'Y';    0  the  angle  between  OX  and  O'X'. 

The  values  of x\  y'  in  terms  of x,  y,  x0,  y0  and  6  are 

x   =  X0  -f.tr  cos  6  —  y  sin  6.  .  .      (1) 

y'  =  y0  -j-  x  sin  6  -\-  y  cos  6.  .         .     (2) 

Differentiating  with  respect  to  the  time, 

*' ==  I  £9  —  (^  sin  ^  -J-  y  cos  6  )  -j-     -f-     *  cos  0  ~  y  s'm  0    ,   (3) 

y '  —  L^o  +  (*  cos  6  —  y  sin  6  )  -  -     -f-      x  sin  0  -f  y  cos  6    .   (4) 
A  second  differentiation  gives 

x0  —  (x  cos  6  —  y  sin  6  )  I  -^-  J  —  (x  sin  6  -\~y  cos  6  ) -j— 

—  2I   (xsm6+yco*0)~     +  I  ^  cos  6>  -  >  sin  ^J  ,      (5) 


RELATIVE   MOTION. 


445 


y 


•,f>. 


( x  sin  6  -f-  y  cos  0  )  (  -jj  )  +  O  cos  0 


\^/ 


') 


£0 


g 


+    2 


[> 


cos  6  —  y  sin 


61 )  ^~|  +  f^  sin  $  +  j;  cos  0~| 


(6) 


O' 


Let  z;  and  /  represent  the  velocity  and  acceleration  of  the  particle 
referred  to  the  base  M,  and  v\  p'  the  velocity  and  acceleration  re- 
ferred to  the  base  M' . 

Then  v  is  equivalent  to  components  x,  y  in  directions  OX,  O  Y; 
and  /  to   components   x,  y  in 
these  directions.  V 

Also  v'  is  equivalent  to 
components  x',  y'  in  directions 
O'X',  O'Y';  and/' to  compo- 
nents x\  y   in  these  directions. 

If  v  and  v'  were  equal  in 
magnitude  and  direction,  the 
value  of  x'  would  equal  the 
sum  of  the  components  of  x 
and  y  in  the  direction  0'X\ 
being  therefore  given  by  the 
last  two  terms  of  equation  (3); 
while    the    value  of  y'    would 

be  given  by  the  last  two  terms  of  equation  (4).  The  remaining 
terms  in  the  values  of  x '  and  y '  as  given  by  these  equations  there- 
fore represent  the  difference  between  v  and  v' \  that  is,  they  repre- 
sent the  components  in  directions  O'X'  and  O'Y'  of  the  vector 
which,  added  to  v,  gives  v' . 

Similiarly,  the  component  of p  in  the  direction  O'X'  is  given  by 
the  last  two  terms  in  equation  (5),  while  its  component  in  the  direc- 
tion O'Y'  is  given  by  the  last  two  terms  of  (6).  The  remaining 
terms  in  the  second  members  of  these  equations,  therefore,  represent 
the  difference  between/  and/';  that  is,  they  represent  the  compon- 
ents in  the  directions  O'X'  and  O'Y'  of  the  vector  which  added  to 
/gives/'. 

It  remains  to  interpret  these  results. 

Interpretation  of  result  in  case  of  velocities.  —  The  first  three 
terms  in  the  second  member  of  equation  (3)  give  the  values  ofir'  for 
a  particle  fixed  to  the  base  M,  as  is  seen  by  the  fact  that  they  are 


Fig.  200. 


446  THEORETICAL    MECHANICS. 

the  terms  which  remain  when  x  and  y  are  constant.  Similarly  for 
the  first  three  terms  in  the  value  of  y  .  Hence  it  is  seen  that  the 
vector  which  represents  the  difference  between  v  and  v '  is  equal  to 
the  velocity  with  respect  to  the  base  M'  of  that  point  of  M  which 
momentarily  coincides  with  the  particle.  We  have  therefore  the 
general  result  that 

The  velocity  of  a  particle  referred  to  any  base  M'  is  equal  to 
the  vector  sum  of  its  velocity  referred  to  another  base  M  and  the 
velocity  (referred  to  M')  of  that  point  of  M  which  momentarily 
coincides  with  the  particle. 

Interpretation  of  result  in  case  of  acceleration. — The  result  is 
less  simple  in  the  case  of  accelerations. 

Notice  first  that  the  acceleration,  with  respect  to  the  primary 
base  M\  of  that  point  of  the  secondary  base  M  which  is  momentar- 
ily the  position  of  the  particle,  is  given  by  the  terms  remaining  in 
the  values  of  x'  and  j/'  when  x  and y  are  constant;  that  is,  the  first 
five  terms  in  the  second  member  of  (5)  represent  its  component  in 
the  direction  O'X',  and  the  corresponding  terms  in  (6)  its  compon- 
ent in  the  direction   O'Y'. 

Again,  the  last  two  terms  in  (5),  and  the  corresponding  terms  in 
(6),  represent  the  resolved  parts  of  p  in  the  directions  OX',  O'Y' 
respectively. 

There  remain  to  be  interpreted  two  terms  in  each  of  the  two 
equations.  In  other  words,  we  have  to  consider  what  vector  it  is 
whose  resolved  parts  are 

7/3 

—  2 -7- (i:  sin  6  -\-  y  cos  ^  in  direction  O'X',  ) 

i  (7) 

_|_2^(^cos  6—y  sin  0)   "         "        O'Y'.) 
dt 

Now         ircos  6  —  y  sin  0  =  resolved  part  of  v  in  direction  OX', 

xsmd  +  ycos6=        "  "     "  v  "        "         O'Y'; 

hence  the  quantities  (7)  are  the  components  oi  a  vector  per- 
pendicular to  v,  of  magnitude  2v{dQ\df)  =  2Vco,  if  co  denotes  the 
angular  velocity  of  the  base  M  with  respect  to  M '. 

It  is  thus  seen  that  equations  (5)  and  (6)  are  equivalent  to  the 
following  proposition  : 


RELATIVE    MOTION.  447 

The  acceleration  of  a  particle  relative  to  any  base  M'  is  equal  to 
the  vector  sum  of  three  components:  (/)  its  acceleration  relative  to 
another  base  M  having  any  plane  motion  with  respect  to  M\  (2)  the 
acceleration  relative  to  M'  of  that  point  of  M  which  momentarily  coin- 
cides with  the  particle,  and  (j)  an  acceleration  directed  at  right 
angles  to  the  velocity  of  the  particle  relative  to  My  of  magnitude 
equal  to  twice  the  product  of  that  velocity  into  the  angular  velocity 
of  M  with  respect  to  M' .  The  direction  of  this  third  component  is 
that  assumed  by  the  vector  representing  the  velocity  relative  to  M  if 
turned  through  go°  in  the  direction  in  which  M  rotates  relatively  to 
M'. 

The  propositions  just  proved  may  be  concisely  expressed  by 
vector  equations.  Thus,  if  u  denotes  the  velocity  relative  to  M'  of 
that  point  of  M  which  instantaneously  coincides  with  the  particle, 
the  relation  between  velocities  is 

v'  =  v  +  u.       .         .         .         .     (1'") 

And  if  the  acceleration  components  numbered  (2)  and  (3)  in  the 
above  statement  be  represented  by  vector  symbols  p"  and/'"  respec- 
tively, the  relation  between  accelerations  is 

>'■.«-/+>"+/".  •     •     •  (3'") 

These  are  seen  to  include  equations  (1")  and  (3")  of  Art.  525. 

Examples. 

1.  Show  that  the  above  general  relation  between  accelerations 
reduces  to  that  given  in  Art.  524  when  the  relative  motion  of  the 
two  bases  is  a  translation  with  uniform  velocity. 

2.  Apply  the  general  result  to  the  case  in  which  the  relative 
motion  of  the  two  bases  is  a  uniform  rotation  about  an  axis  fixed  in 
both,  and  compare  with  the  result  reached  in  Art.  525. 

3.  In  what  cases  does  the  component/'"  vanish  ? 

4.  A  wheel  of  radius  a  rolls  uniformly  along  the  ground,  the 
velocity  of  its  center  being  V.  Determine  the  velocity  and  acceler- 
ation relative  to  the  earth  of  a  point  in  the  circumference  {a)  when 
in  its  highest  position,  (J?)  when  in  its  lowest  position. 

5.  In  Ex.  4,  suppose  a  particle  to  move  uniformly  along  a  diam- 
eter of  the  wheel,  and  that  its  relative  path  is  vertical  when  it  passes 
through  the  center.  Determine  its  velocity  and  acceleration  relative 
to  the  earth  at  that  instant. 


448  THEORETICAL    MECHANICS. 

527.  Three-Dimensional  Motion  of  a  Particle. — The  foregoing 
discussion  assumes  the  particle  to  move  in  a  plane  parallel  to  the 
relative  motion  of  the  bases.  In  order  to  remove  this  restriction, 
let  the  point  A  (Fig.  200)  represent  the  orthographic  projection  of 
the  position  of  the  particle  upon  a  plane  parallel  to  the  relative  motion 
of  the  bases.     Reasoning  as  in  Art.  525,  it  is  seen  that 

When  the  motion  of  the  particle  is  unrestricted,  the  relation  be- 
tween velocities  is  the  same  as  in  the  case  of  plane  motion;  while  tlie 
relation  between  accelerations  is  changed  by  the  substitution  for  the 
velocity  relative  to  M  of  the  resolved  part  of  that  velocity  parallel  to 
the  plane  of  the  relative  motion  of  the  bases.  This  change  affects 
only  the  component p" ' . 

Example. 

Show  that  the  above  proposition  holds  when  the  relative  motion 
of  the  bases  is  equivalent  to  a  plane  motion  combined  with  any 
translatory  motion. 

§  3.   Dynamics  of  Relative  Motion. 

528.  Equation  of  Motion  for  Any  Base. — The  importance  of 
the  choice  of  a  base  of  reference  for  the  purpose  of  applying  the 
laws  of  motion  was  shown  in  Art.  518.  This  is  made  more  definite 
by  the  foregoing  analysis.     If  the  equation 

force  ==  mass  X  acceleration 

holds  when  the  motion  of  a  particle  is  referred  to  a  base  M' ,  it  will 
not  hold  when  the  motion  is  referred  to  another  base  M,  unless  the 
acceleration  has  the  same  value  for  the  two  bases  ;  and  this  will  be 
true  only  if  the  motion  of  M  relative  to  M'  is  a  translation  with 
uniform  rectilinear  velocity. 

It  is,  however,  easy  to  write  a  dynamical  equation  which  will  hold 
for  any  base  which  itself  moves  in  a  known  manner  with  respect  to 
an  ultimate  base.  Thus,  applying  the  result  of  Art.  526,  let  the 
base  M'  be  an  ultimate  base,  and  let  P'  denote  the  resultant  of  all 
forces  acting  upon  the  particle,  the  remaining  notation  being  as 
before.     The  general  equation  of  motion  is 

P'  =  mp'  5  .     (1) 

or,  by  equation  (3'")  of  Art.  526,  we  may  write  the  vector  equation 

P'  =  mp  -j-  mp"  -f-  rnp'" .       .         .         .     (2) 


RELATIVE   MOTION.  449 

This  equation  serves  to  determine/  when  the  forces  are  known,  if 
the  motion  of  M  relative  to  M'  is  known  so  that  /  "  and  p '"  can  be 
computed. 

529.  Fictitious  Forces — The  last  equation  may  be  written 

P'  —  mp"  —  mp'"  =  mp.     .         .         .     (3) 
Or,  if  P  is  written  for  the  vector  P'  —  mp"  —  mp"\ 

P=mp (4) 

The  equation  of  motion  referred  to  the  secondary  base  M  thus  has 
the  same  form  as  that  referred  to  an  ultimate  base,  P  taking  the 
place  of  the  ' '  resultant  force ' '  acting  upon  the  particle.  But  the 
forces  of  which  P  is  the  resultant  include,  besides  the  actual  forces 
(whose  resultant  has  been  called  P'),  two  "fictitious"  forces  whose 
values  ( — mp"  and  — mp'")  depend  upon  the  motion  of  the  second- 
ary base  with  respect  to  an  ultimate  base. 

Equations  of  equilibrium.— If  the  particle  is  at  rest  relative  to 
the  base  M,  the  acceleration/'"  is  o,  and  the  only  fictitious  force  to 
be  introduced  into  the  equation  of  equilibrium  is  the  force  — mp  ". 

Examples. 

[In  these  examples  it  may  be  assumed  that  the  earth  is  an  ulti- 
mate base.  The  error  involved  in  this  assumption  will  be  considered 
in  Art.  531.] 

1.  A  ball  is  thrown  back  and  forth  by  two  persons  on  the  deck 
of  a  steamer  which  moves  with  uniform  speed  in  a  constant  direction. 
How  does  the  motion  of  the  boat  affect  the  game  ?  What  is  the 
effect  of  a  change  in  the  speed  of  the  steamer  ? 

2.  An  elevator  platform  starts  from  rest  with  an  upward  accelera- 
tion gj<\,  then  moves  upward  with  constant  velocity,  then  has  down- 
ward acceleration  g/6.  Taking  the  platform  as  base,  write  the 
equation  of  motion  for  a  body  resting  upon  the  platform,  for  each  of 
the  three  periods  of  the  motion. 

3.  An  elevator  platform,  starting  from  rest,  moves  with  a  down- 
ward acceleration  which  increases  uniformly  from  o  to  2g  in  a  period 
of  4  seconds,  (a)  Write  the  equation  of  motion  for  a  body  resting 
on  the  platform,  taking  the  platform  as  base,  (b)  Determine  com- 
pletely the  motion  of  the  body  during  4  seconds.  (c)  How  far  will 
it  be  from  the  platform  at  the  end  of  that  time  ? 


45°  THEORETICAL    MECHANICS. 

4.  A  horizontal  platform  rotates  uniformly  about  a  fixed  vertical 
axis.  Write  the  equations  of  motion  for  a  body  at  rest  upon  the 
platform,  taking  the  latter  as  base. 

Ans.  The  actual  forces  acting  upon  the  body  are  (1)  its  weight 
vertically  downward,  and  (2)  the  supporting  force  exerted  by  the 
platform.  Of  the  fictitious  forces,  one  vanishes,  since  the  velocity 
relative  to  the  platform  is  zero  ;  the  other,  — mp"  y  is  a  force  directed 
at  right  angles  to  the  axis  of  rotation  and  away  from  it.  If  r  is  the 
distance  of  the  particle  from  the  axis  of  rotation  and  co  the  angular 
velocity  of  the  platform,  p  *  =  o>V. 

5.  In  Ex.  2.  of  Art.  525,  assuming  the  earth  to  be  an  ultimate  base, 
(a)  determine  the  resultant  of  the  actual  forces  acting  upon  the  par- 
ticle when  distant  r  from  the  axis,  {b)  What  fictitious  forces  must 
be  assumed  to  act  if  the  platform  is  taken  as  base  ? 

6.  A  particle  slides  in  a  smooth  straight  tube  which  rotates  uni- 
formly about  a  vertical  axis  at  right  angles  to  its  length.  The  actual 
forces  acting  upon  the  particle  being  its  weight  and  the  pres- 
sure of  the  tube,  (a)  write  the  equation  of  motion,  taking  the 
tube  as  base,  (b)  If  the  particle  passes  through  the  axis  of  rotation 
with  velocity  V  relative  to  the  tube,  what  will  be  its  velocity  when 
distant  r  from  this  position  ? 

Ans.   (5)  v2  —  V2  +  ri(°~,  w  being  the  angular  velocity. 

7.  Solve  Ex.  6  assuming  the  tube  to  be  inclined  at  angle  a  to  the 
horizontal,  the  axis  still  being  vertical. 

Ans.   (&)  v*  =  V'1  +  r2co*  cos2  a  -f  2gr  sin  a. 

530.  Centrifugal  Force. — When  the  secondary  base  rotates  uni- 
formly about  a  fixed  axis,  the  fictitious  force  — mp  "  is  directed  away 
from  the  axis  of  rotation,  and  its  magnitude  is  wft>V,  co  being  the 
angular  velocity  and  r  the  distance  of  the  particle  from  the  axis. 
This  is  called  the  centrifugal  force  due  to  the  rotation  of  the  base  to 
which  the  motion  is  referred.* 

Whatever  motion  the  secondary  base  may  have  with  respect  to  an 
ultimate  base,  the  component  of  the  fictitious  force  — mp  "  directed 
away  from  the  instantaneous  axis  may  be  called  the  centrifugal  force. 

531.  The  Earth  as  Base. — The  earth  is  usually  the  most  con- 
venient base  for  estimating  the  motions  of  terrestrial  bodies.  The 
fictitious  forces  which  must  be  introduced  into  the  equations  of  mo- 
tion in  order  to  take  account  of  the  known  motions  of  the  earth  are 
for  many  purposes  negligible  in  comparison  with  the  actual  forces. 

*  This  is  the  proper  use  of  the  term  centrifugal  force,  to  which  allusion 
was  made  in  the  foot-note  on  p.  271. 


RELATIVE    MOTION.  45 1 

The  most  important  of  these  motions  of  the  earth  is  the  diurnal  ro- 
tation. 

Correction  for  earth's  rotation.  —  Let  a  frame  of  reference  be 
taken  consisting  of  three  rectangular  axes  intersecting  at  the  earth' s 
center,  one  coinciding  with  the  axis  of  rotation,  a  second  lying  in 
the  plane  through  that  axis  and  a  fixed  star,  the  third  perpendicular 
to  this  plane.  The  motion  of  the  earth  with  respect  to  this  frame  is 
a  rotation  with  uniform  angular  velocity  a>  =  0.00007292  rad.  per 
sec.  (Ex.  1,  p.  275).  The  effect  of  this  rotation  upon  the  motion  of 
a  particle  relative  to  the  earth  is  to  give  it  two  components  of  accel- 
eration which  are  the  negatives  of p"  and/'"  as  defined  in  Art.  526. 
In  other  words,  the  fictitious  forces  to  be  introduced  are  — mp"  and 
— mp'" .  The  former  of  these  is  the  ''centrifugal"  force  whose 
magnitude  is  mcolry  directed  away  from  the  axis  of  rotation  ;  the 
latter  is  a  force  of  magnitude  2nmv  sin  a,  a  being  the  angle  between 
V  and  the  axis  of  rotation.  This  latter  force  is  always  perpendicular 
to  the  axis  of  rotation  and  to  the  direction  of  the  relative  motion, 
in  such  a  way  that  the  particle  is  deflected  toward  the  right  in  the 
northern  hemisphere  and  toward  the  left'in  the  southern,  as  seen  by 
a  person  standing  upon  the  earth  and  facing  in  the  direction  of  the 
relative  motion  of  the  particle. 

The  greatest  value  of  the  acceleration  p  "  occurs  at  the  equator, 
being  3. 391  C.  G.  S.  units*  or  1/289  °f g-  Practically,  the  acceler- 
ation — p  "  cannot  be  separated  from  the  acceleration  due  to  gravity 
except  by  computation.  The  value  of^*  as  determined  experiment- 
ally at  any  place  is  the  resultant  of  the  component  —p  "  and  the  ac- 
celeration due  to  the  attraction  of  the  earth.  In  practice,  therefore, 
the  centrifugal  force  due  to  the  earth's  rotation  is  always  taken 
account  of  when  the  weight  of  the  body  is  one  of  the  forces  entering 
the  equation  of  motion. 

The  acceleration  p '"  is  small  unless  the  velocity  of  the  particle 
relative  to  the  earth  is  very  great.  If  v  sin  a  =  100  meters  per 
second,/'"  =  1.458  C.  G.  S.  units. 

The  deflection  of  a  particle  toward  the  right  in  the  northern 
hemisphere  and  toward  the  left  in  the  southern  has  an  important 
effect  upon  the  circulation  of  the  atmosphere.  It  is  this  which  de- 
termines the  direction  of  rotation  of  cyclonic  disturbances. 

*See  Ex.  3,  p.  275. 


452 


THEORETICAL   MECHANICS. 


Correction  for  motions  of  the  earth  about  the  sun  and  the  moon. 
— The  origin  of  the  frame  of  reference  described  at  the  beginning  of 
this  Article  is  at  the  center  of  the  earth,  and  the  frame  has  therefore 
a  motion  of  translation  due  to  the  orbital  motion  of  the  earth's  center 
of  mass.  By  reason  of  the  attraction  of  the  sun,  the  center  of  mass 
of  the  earth  describes  an  approximately  circular  orbit  about  the  com- 
mon center  of  mass  of  sun  and  earth  ;  and  by  reason  of  the 
moon's  attraction  it  describes  an  approximately  circular  orbit  about 
the  common  center  of  mass  of  moon  and  earth.  These  are  most 
simply  considered  separately.  The  fictitious  forces  which  must  be 
assumed  to  act  upon  any  terrestrial  particle  because  of  these  motions, 
if  the  frame  of  reference  above  described  be  taken  as  base,  are  easily 
computed  from  the  general  principle  of  Art.  524.  The  most  im- 
portant application  in  which  these  corrections  need  to  be  made  is  in 
the  theory  of  the  tides. 


INDEX. 


Acceleration,  172,  173,  212,  214. 
average,  213. 

axial  components  of,  239. 

formula  for,  173. 

polar  resolution  of,  241. 

tangential  and  normal  compo- 
nents of,  244. 
Action  at  a  distance,  19. 
Activity,  316. 

Angular  impulse,  292,  383. 
Angular  momentum,  287,  382. 

principle  of,  294,  326. 

value  of,  387,  388,  395. 
Angular  motion  of  material  system, 
325,  326. 

of  particle,  281. 

of  rigid  body,  363,  372. 
Attraction  between  particles,  155. 

between  spheres,  164. 

of  sphere  upon  particle,  162. 

of  spherical  shell  upon  exterior 
particle,  160. 

of  spherical  shell  upon  interior 
particle,  158,  162. 
Balance,  86. 
Base  for  specifying  motion,  432,  433. 

ultimate,  434. 
Body,  2,  228. 
Catenary,  118. 
Center  of  mass,  127,  129. 
Central   force,  motion  under,  193- 
198,  253,  260. 

varying  as  distance,  253. 

varying  inversely  as  square  of 
distance,  258. 
Centrifugal  force,  271,  450. 
Centroid,  124-140. 

determined  by  integration,  135- 
140. 

of  parallel  forces,  125,  126. 

of  plane  area,  131-134. 

of  volume,  surface,  line,  130. 


Circular  motion,  uniform,  270. 
Collision  of  bodies,  282,  340. 
Concurrent  forces,  composition  and 
resolution  of,  27-35,  14 1»  142. 
equilibrium  of,  36-48,  143. 
Configuration  defined,  407. 

energy  of,  408. 
Connected  particles,  motion  of,  203. 
Conservation  of  energy,  417. 
Conservative  system,  410. 
law  of  force  in,  410. 
principle  of  work  and  energy 
for,  412. 
Constant  force,  motion  under,  188, 

249,  252. 
Constant  of  gravitation,  164,  166. 
Constrained  motion,  260-276. 
Continuity  of  matter,  3,  228. 
Coordinates  of  position  of  particle, 
236,  321. 
of  rigid  body,  345,  360. 
Coplanar  forces,   composition  and 
resolution  of,  60-64. 
equilibrium  of,  65-91. 
Cord,  flexible,  22,  1 14-123. 
Couples,   composition  of,  59,  145, 
146. 
coplanar,  56-60. 
impulsive,  399. 
in  three  dimensions,  144-146. 
representation  of  by  vector,  144. 
Curved  path,  motion  in,  208-236. 
D'Alembert's    principle,   327,   329, 

37i. 
Degrees  of  freedom,  346. 
Density,  128. 

Dimensional  equations,  7. 
Dimensions  of  derived  units,  6, 157, 

184,  280,  300,  310. 
Displacement,  167. 

given  by  vector,  208. 
of  rigid  body,  360-361. 


454 


INDEX. 


Dynamics,  4. 

of  relative  motion,  448. 
Earth  as  base  for  specifying  motion, 

45o. 
Effective  forces,  327,  370. 

moments  of,  350. 

resultant  of,  357 . 
Efficiency,  419. 
Energy,  308,  312,  407. 

a  definite  quantity,  409. 

and  heat,  equivalence  of,  414. 

forms  of,  415. 

of  a  particle,  307-311. 

of  a  material  system,  311-317, 
407-412. 

of  a  rigid  system,  419. 

transfer  of,  412. 

utilized  and  lost,  417. 
Equilibrium,  conditions  of  for  con- 
current forces,  36-38,  143. 

for  coplanar  forces,  65-68. 

for  non-coplanar   forces,    149- 

151. 

definition  of,  26. 

of  flexible  cords,  1 14-123. 

of  parts  of  bodies,  92. 

of  system  of  bodies,  99. 

work-condition  of,  318,  424. 
Flexible  cords,  equilibrium  of,  114- 

123. 
Force,  conception  of,  2,  16-18,  177. 

defined  by  laws  of  motion,  225- 
228. 

effect  of  constant,  177,  184,  217. 

effect  of  variable,  185,  220. 

gravitation  unit  of,  23. 

kinetic  or  absolute  unit  of,  222. 
Forces,  classes  of,  18-22. 

equal,  defined  by  law  of  action 
and  reacti  ,  227. 

external  and  internal,  92. 

fictitious,  449. 
Freedom,  degrees  of,  346. 
Friction,  106-113. 

angle  of,  108. 

coefficient  of,  108. 


Friction,  in  machines,  11 1. 

laws  of,  107. 
Gram,  24. 

Gravitation,  constant  of,    156-164, 
166, 

law  of,  155,  158. 
Gravitation  system  of  units,  22-25. 

unit  of  mass,  156,  165,  166. 
Gravity,  motion  under,  189, 199, 249. 
Harmonic  motion,- 196. 
Heat  equivalent  to  energy,  414. 
Hodograph,  211,  239. 
Homogeneous  body,  128,  129. 
Impact,  282-284. 
Impulse,  278,  289,  291,  383. 

and  momentum,  equations  of, 
for  material  system,  384, 
392. 
for  particle,  279,  290,  291. 
for  rigid  body,  388,  395. 

angular,  392. 

instantaneous,  396,  400. 

resultant,  292. 

sudden,  280,  281,  296,  385. 
Impulsive  couple,  399. 
Inertia-ellipse,  344. 
Instantaneous  motion  of  rigid  body, 

362. 
Jointed  frame,  95,  96. 
Kinematics,  4. 
Kinetics,  5. 

Kinetic  energy  of  material  system, 
407. 

of  particle,  308,  309. 
Kinetic  systems  of  units,  181-184. 
Lever,  83. 

Machine,  82,  316,  426. 
Mass,  2. 

defined  by  laws  of  motion,  225, 
226. 
Mass-center,   motion   of,  322,   324, 

348,  352. 
Material  body,  2. 

system,  321. 
Mechanics  defined,  1. 

subdivisions  of,  4. 


INDEX. 


455 


Metric  system,  24. 
Moment  with  respect  to  an  axis,  35, 
150. 

with  respect  to  a  plane,  130. 

with  respect  to  a  point,  13,  35. 
Moment  of  inertia  of  plane  area, 
339-345- 

of  rigid  body,  33i"339- 
Moments,  theorem  of,  14,  54-56. 
Momentum,  277,  287,  291,  382. 

acceleration  of,  277,  288. 

angular,  287. 

linear,  386,  387. 

resultant,  394,  395. 
Motion,  absolute  and  relative,  436. 

laws  of,  177,  223. 
Newton's  first  law,  16,  177. 

third  law,  17. 

three  laws,  223. 
Parallel  forces,    resultant   of    two, 

5i 
Parallelogram  of  forces,  28,  221. 
Parallelopiped  of  forces,  141. 
Particle,  2,  228. 
Passive  resistances,  19. 
Pendulum,  compound,  354. 

simple,  266. 
Percussion,  center  of,  390. 
Plane    motion     of     particle,    236- 
276. 

of  rigid  body,  360-381. 
Plane  motions  of  rigid  body,  com- 
position and    resolution    of, 
364-370. 
Polygon  of  forces,  30. 
Position,  3. 

in  straight  line,  167. 

given  by  vector,  208. 
Potential  energy,  312,  408. 
Pound  force,  23. 
Pound  mass,  23 
Power,  316. 

Product  of  inertia,  33S,  341. 
Projectile,  249. 
Pulley,  83,  84,  in. 
Pulleys,  system  of,  85,  430. 


Quantities,    numerical    representa- 
tion of,  5. 
Quantity,  fundamental  kinds  of,  4. 
Radius  of  gyration,  332,  344. 
Relativity  of  motion,  232. 
Resisting  medium,  motion  in,  203. 
Resolution  of  forces,  26,  31,  50, 141. 
Restitution,  coefficient  of,  285. 
Resultant,  definition  of,  26. 

of  concurrent  forces,  27-30, 142. 
of  coplanar  forces,  60,  64. 
of  couples,  59,  145,  146. 
of  non -coplanar  forces,  142, 149. 
Rigid  body,  3,  49. 

motion  of,  346-381 . 
Rigid  system,  energy  of,  419. 
Rotating  body,  equations  of  motion 

of,  353- 
Rotation,  347,  349-359. 

of  earth,  its  influence  on  ap- 
parent weight,  273. 
Scalar,  8. 

Simultaneous  motions,  229-235. 
Smooth  surface,  21 
Space,  436. 
Standard  condition  for  estimating 

energy,  313,  408. 
Statics,  5. 
Statics  a  case  of  kinetics,  252,  379- 

381. 
Strain,  104. 
Stress  defined,  17. 
work  of,  405. 
Stresses,  external  and  internal,  102- 

104. 
Symmetry,  131. 
Time,  435,  436. 
Translation,  347. 
Triangle  of  forces,  28. 
Units,  fundamental  and  derived,  6. 
Vector,  7. 

addition,  8. 

increment  of  variable,  1 1 . 
localized,  13. 
resolution  of,  12. 
subtraction,  9. 


456 


INDEX. 


Velocity,  167,  209. 

average,  169,  209. 

axial  components  of,  238. 

formula  for,  170,  210. 

polar  resolution  of,  240. 

sudden  change  of,  215. 

tangential    and    normal    com- 
ponents of,  244. 
Velocity-increment,  171,  212. 
Virtual  work,  principle  of,  318-320, 

424,  425. 
Vis  viva,  306. 


Weight,  23. 

influence  of  earth's  rotation  on, 

273- 
Work  and  energy,  principle  of,  309, 

314,  412,  420. 
Work  and  vis  viva,  equation  of,  306. 
Work  done  by  body,  307. 

done  by  force,  298,  299,  301. 

external  and  internal,  407. 

of  central  force,  303. 

of  stress,  405. 

virtual,  318,  424. 


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